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Finding Taylor or Maclaurin Series for a Function

Build Taylor or Maclaurin series and justify function representation with a vanishing remainder.

College Board Topic 10.14 · CED p. 194 (PDF p. 199) · CED effective Fall 2020; project course outline 2026–27.

The Taylor series at a extends the polynomial formula indefinitely: ∑n=0∞f(n)(a)n!(x-a)n. Its partial sums are Taylor polynomials. Having infinitely many derivatives lets you write coefficients, but equality to f on an interval also requires the remainder to tend to zero there. A series that converges need not automatically equal the function whose derivatives generated it.

Four Maclaurin foundations do most of the work: 1/(1-x)=∑n=0∞xn for |x|<1; ex=∑n=0∞xn/n! for all real x; sinx=∑n=0∞(-1)nx2n+1/(2n+1)! for all real x; and cosx=∑n=0∞(-1)nx2n/(2n)! for all real x. Their coefficient patterns come from derivatives at zero or the geometric identity.

For eˣ, sin x and cos x on any fixed bounded interval, a suitable derivative bound divided by the factorial makes the Taylor remainder approach zero. That establishes equality rather than just a formal coefficient pattern. In sine and cosine, parity matters: odd powers for sine and even powers for cosine.

A Taylor series can have a different center. For ln x about a=1, use the powers (x−1)ⁿ and coefficients (−1)ⁿ⁺¹/n for n≥1. The series represents ln x on 0<x≤2; checking x=2 gives alternating convergence, while x=0 diverges. Interpretation includes the center, domain of representation and partial-sum approximation, not only copying coefficients.

Let's work through it

  1. For cos(2x), start with the known cosine series. Why: A foundation series avoids recalculating all derivatives.
  2. Substitute 2x to get ∑n=0∞(-1)n(2x)2n/(2n)!. Why: The substitution goes into the whole power, not just one coefficient.
  3. The first three nonzero terms are 1-2x2+(2/3)x4, and the series represents cos(2x) for all real x. Why: The cosine series has infinite radius; its Taylor polynomial is the corresponding partial sum.

Establish the foundation before substituting

The Maclaurin series for 1/(1−x) is the geometric series 1+x+x²+x³+⋯=Σₙ₌₀^∞ xⁿ, with first term 1, common ratio x, and center 0. The equality holds for |x|<1. To see why, the finite sum S_N=1+x+⋯+xᴺ satisfies (1−x)S_N=1−x^(N+1), so S_N=(1−x^(N+1))/(1−x) when x≠1. For |x|<1 the remainder x^(N+1)/(1−x) tends to zero, giving the stated function identity.

The endpoints are excluded: at x=1 the terms stay 1, and at x=−1 they alternate between 1 and −1 without tending to zero. The ratio-test value 1 at a boundary does not establish convergence. For |x|>1, terms grow in magnitude. A finite rational value at some outside point is not evidence that this Maclaurin series converges there.

Only after writing the foundation should you replace its variable. Set u=−x²/2 and multiply by 1/2 to get 1/(2+x²)=(1/2)Σₙ₌₀^∞ (−x²/2)ⁿ=Σₙ₌₀^∞ (−1)ⁿx^(2n)/2^(n+1). The condition is |−x²/2|<1, so |x|<√2. At either endpoint the terms are (−1)ⁿ/2 and do not approach zero. The substituted power series therefore has radius √2 and interval (−√2,√2), even though the rational function is defined for every real x.

A top strip shows the foundation ratio u with the strict geometric domain −1<u<1 and open endpoint circles. The substitution u=−x²/2 maps the derived series to −√2<x<√2. A lower table compares x=0, 1/2, and 2: ratios 0, −1/8, and −2; the first two converge and the last diverges even though 1/(2+x²) is finite.
Start with 1/(1−u)=Σ uⁿ for |u|<1, then substitute the entire expression −x²/2. The rational function exists outside the series interval; its existence does not make that particular series converge.
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A top strip shows the foundation ratio u with the strict geometric domain −1<u<1 and open endpoint circles. The substitution u=−x²/2 maps the derived series to −√2<x<√2. A lower table compares x=0, 1/2, and 2: ratios 0, −1/8, and −2; the first two converge and the last diverges even though 1/(2+x²) is finite.

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LIM-8.F.1

Substitute into every sine power

The sine foundation is sin u=Σₙ₌₀^∞ (−1)ⁿu^(2n+1)/(2n+1)! for every real u, with angles measured in radians. Substitute the whole expression 2x into every power: sin(2x)=Σₙ₌₀^∞ (−1)ⁿ2^(2n+1)x^(2n+1)/(2n+1)!. The factor 2 is raised to the same power as x; replacing every coefficient by just twice its old value is incorrect.

For x≠0, division by x yields sin(2x)/x=Σₙ₌₀^∞ (−1)ⁿ2^(2n+1)x^(2n)/(2n+1)!. This is an even-power series: 2−(4/3)x²+(4/15)x⁴−(8/315)x⁶+⋯. The original quotient is undefined at zero, but its limit is 2. Defining that value makes the function continuous there, and the series’s constant term agrees. The series represents this extended function for every real x, with infinite radius.

A polynomial such as P₄=2−(4/3)x²+(4/15)x⁴ is a partial sum of the infinite series. It is an approximation away from zero, not the entire function. On |x|≤1/2 the alternating magnitudes decrease because each successive ratio is 4x²/((2n+2)(2n+3))≤1/6. Thus the first omitted nonzero term gives |S(x)−P₄(x)|≤8|x|⁶/315. The represented infinite series and the finite approximation need different equality and approximation symbols.

An error plot over −0.5≤x≤0.5 compares the actual nonnegative difference P₄(x)−sin(2x)/x, using the continuous value 2 at zero, with the upper bound 8|x|⁶/315. Both vanish at zero; the bound lies just above the actual error. A labeled table at x=0.25 gives approximation 1.917708333, independent value 1.917702154, observed error about 6.179×10⁻⁶, and guaranteed bound about 6.200×10⁻⁶.
The sine foundation yields the full infinite series; P₄ is only its first three nonzero terms. The error curve illustrates a checked case, while decreasing alternating magnitudes justify the bound on this stated interval.
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An error plot over −0.5≤x≤0.5 compares the actual nonnegative difference P₄(x)−sin(2x)/x, using the continuous value 2 at zero, with the upper bound 8|x|⁶/315. Both vanish at zero; the bound lies just above the actual error. A labeled table at x=0.25 gives approximation 1.917708333, independent value 1.917702154, observed error about 6.179×10⁻⁶, and guaranteed bound about 6.200×10⁻⁶.

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LIM-8.F.2LIM-8.E.1

Integrate a substituted exponential series

The exponential foundation is eᵘ=Σₙ₌₀^∞ uⁿ/n! for every real u. Substituting u=−t² gives e^(−t²)=Σₙ₌₀^∞ (−1)ⁿt^(2n)/n!=1−t²+t⁴/2−t⁶/6+⋯. The minus sign and square affect every indexed term; the factorial still belongs to n, not to the resulting power 2n.

Define W(x)=∫₀ˣ e^(−t²)dt. Since the substituted power series has infinite radius, termwise integration is valid on every finite interval from 0 to x. Therefore W(x)=Σₙ₌₀^∞ (−1)ⁿx^(2n+1)/(n!(2n+1))=x−x³/3+x⁵/10−x⁷/42+⋯ for all real x. Both the integral and series vanish at zero, so no undetermined constant remains. Differentiating the constructed series recovers the substituted exponential series, checking the denominators and powers.

The integrand is positive, so W is increasing. Its even integrand makes W odd: W(−x)=−W(x). At a chosen point, a finite polynomial approximates the integral; an alternating-series bound controls the unseen tail when its magnitudes decrease to zero. For |x|≤1/2, successive integrated magnitudes have ratio x²(2n+1)/((n+1)(2n+3))<1. Thus P₅=x−x³/3+x⁵/10 has error at most |x|⁷/42 on this stated interval.

The graph of y=e^(−t²) on 0≤t≤1 is positive and decreasing. Shading under it from t=0 to 0.5 represents W(0.5)=∫₀^0.5 e^(−t²)dt. A separate numerical panel gives P₅(0.5)=443/960≈0.461458333, independent integral≈0.461281006, and error at most 1/5376≈0.000186012.
Substitute −t² into the exponential foundation before integrating. The area is the function W, while the polynomial is an approximation to it; W(0)=0 fixes the integration constant.
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The graph of y=e^(−t²) on 0≤t≤1 is positive and decreasing. Shading under it from t=0 to 0.5 represents W(0.5)=∫₀^0.5 e^(−t²)dt. A separate numerical panel gives P₅(0.5)=443/960≈0.461458333, independent integral≈0.461281006, and error at most 1/5376≈0.000186012.

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LIM-8.F.2

The function exists where this series diverges

Represent f(x)=1/(2+x²) using the geometric Maclaurin foundation. At x=1/2, approximate with the first three nonzero terms and compare a guaranteed bound with the actual error.

A top strip shows the foundation ratio u with the strict geometric domain −1<u<1 and open endpoint circles. The substitution u=−x²/2 maps the derived series to −√2<x<√2. A lower table compares x=0, 1/2, and 2: ratios 0, −1/8, and −2; the first two converge and the last diverges even though 1/(2+x²) is finite.
Start with 1/(1−u)=Σ uⁿ for |u|<1, then substitute the entire expression −x²/2. The rational function exists outside the series interval; its existence does not make that particular series converge.
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A top strip shows the foundation ratio u with the strict geometric domain −1<u<1 and open endpoint circles. The substitution u=−x²/2 maps the derived series to −√2<x<√2. A lower table compares x=0, 1/2, and 2: ratios 0, −1/8, and −2; the first two converge and the last diverges even though 1/(2+x²) is finite.

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  1. Start explicitly with 1/(1−u)=Σₙ₌₀^∞ uⁿ for |u|<1. Substitute u=−x²/2 and multiply by 1/2: f(x)=Σₙ₌₀^∞ (−1)ⁿx^(2n)/2^(n+1).
  2. The strict condition gives |x|<√2. At x=±√2, the terms (−1)ⁿ/2 fail to tend to zero, so neither endpoint is included. At x=2 the ratio is −2 and the terms also fail to tend to zero, although f(2)=1/6.
  3. The first three nonzero terms give P₄(x)=1/2−x²/4+x⁴/8. At x=1/2, P₄=57/128=0.4453125.
  4. The alternating terms decrease at this point, so the error is at most (1/2)⁶/16=1/1024≈0.000976563. The exact value is 4/9, so the actual error is 57/128−4/9=1/1152≈0.000868056, smaller than the bound.

Answer and check

The infinite series represents f on (−√2,√2), with radius √2. At 1/2, f≈0.4453125 and the guaranteed error is at most 1/1024. Function domain and series convergence interval are different objects.

LIM-8.F.1

An infinite sine construction with a removable point

Let S(x)=sin(2x)/x for x≠0 and S(0)=2. Use the sine foundation to construct its Maclaurin series and estimate S(1/4) with the first three nonzero terms. A supplied independent value is S(1/4)≈1.917702154.

An error plot over −0.5≤x≤0.5 compares the actual nonnegative difference P₄(x)−sin(2x)/x, using the continuous value 2 at zero, with the upper bound 8|x|⁶/315. Both vanish at zero; the bound lies just above the actual error. A labeled table at x=0.25 gives approximation 1.917708333, independent value 1.917702154, observed error about 6.179×10⁻⁶, and guaranteed bound about 6.200×10⁻⁶.
The sine foundation yields the full infinite series; P₄ is only its first three nonzero terms. The error curve illustrates a checked case, while decreasing alternating magnitudes justify the bound on this stated interval.
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The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.

An error plot over −0.5≤x≤0.5 compares the actual nonnegative difference P₄(x)−sin(2x)/x, using the continuous value 2 at zero, with the upper bound 8|x|⁶/315. Both vanish at zero; the bound lies just above the actual error. A labeled table at x=0.25 gives approximation 1.917708333, independent value 1.917702154, observed error about 6.179×10⁻⁶, and guaranteed bound about 6.200×10⁻⁶.

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  1. Substitute 2x into sin u=Σₙ₌₀^∞ (−1)ⁿu^(2n+1)/(2n+1)!, then divide each term by x for x≠0. The result is Σₙ₌₀^∞ (−1)ⁿ2^(2n+1)x^(2n)/(2n+1)!.
  2. At zero the series equals its constant term 2, agreeing with the given continuous extension. The foundation represents sine for every real argument, so this new series represents S for all real x.
  3. P₄(1/4)=2−1/12+1/960=1841/960≈1.917708333. This is a partial sum, not an exact S value.
  4. At x=1/4 the next magnitude is 8(1/4)⁶/315=1/161280≈0.0000062004. Successive magnitudes decrease to zero, so the alternating bound applies. The supplied check gives an observed difference about 0.0000061793, within the bound.

Answer and check

S(x)=Σₙ₌₀^∞ (−1)ⁿ2^(2n+1)x^(2n)/(2n+1)! for all real x; S(1/4)≈1841/960 with error at most 1/161280. The signed first omitted term is negative, so P₄ is an upper estimate at this positive point.

LIM-8.F.2

A series for a definite integral

Let W(x)=∫₀ˣ e^(−t²)dt. Construct its Maclaurin series from the exponential foundation, then use the first three nonzero terms at x=1/2. A supplied numerical integral is 0.461281006.

The graph of y=e^(−t²) on 0≤t≤1 is positive and decreasing. Shading under it from t=0 to 0.5 represents W(0.5)=∫₀^0.5 e^(−t²)dt. A separate numerical panel gives P₅(0.5)=443/960≈0.461458333, independent integral≈0.461281006, and error at most 1/5376≈0.000186012.
Substitute −t² into the exponential foundation before integrating. The area is the function W, while the polynomial is an approximation to it; W(0)=0 fixes the integration constant.
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The graph of y=e^(−t²) on 0≤t≤1 is positive and decreasing. Shading under it from t=0 to 0.5 represents W(0.5)=∫₀^0.5 e^(−t²)dt. A separate numerical panel gives P₅(0.5)=443/960≈0.461458333, independent integral≈0.461281006, and error at most 1/5376≈0.000186012.

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  1. Start from eᵘ=Σₙ₌₀^∞ uⁿ/n! for all real u and substitute u=−t². Integrate each (−1)ⁿt^(2n)/n! from 0 to x.
  2. W(x)=Σₙ₌₀^∞ (−1)ⁿx^(2n+1)/(n!(2n+1)). The radius is infinite, and the definite integral fixes W(0)=0.
  3. P₅(1/2)=1/2−1/24+1/320=443/960≈0.461458333. The next omitted magnitude is (1/2)⁷/42=1/5376≈0.000186012.
  4. At this point the integrated magnitudes decrease to zero, so the bound applies. The independent integral gives an observed difference about 0.000177327, below 1/5376. The next term is negative, so P₅ is an upper estimate.

Answer and check

W(x)=Σₙ₌₀^∞ (−1)ⁿx^(2n+1)/(n!(2n+1)) for all real x. W(1/2)≈443/960 with guaranteed error at most 1/5376. The numerical check illustrates the result; the alternating argument supplies the guarantee.

LIM-8.F.2

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Question 1

What is the constant term in the Maclaurin series for eˣ?

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Retrieval 1

What is the constant term in the Maclaurin series for eˣ?
Worked answer and reason
1

It is e⁰/0!.

Retrieval 2

Does the sine Maclaurin series contain a nonzero x² term?
Worked answer and reason
No.

Even derivatives of sine at zero vanish.

Catch the tempting mistake

Every infinitely differentiable function equals its Taylor series.
The remainder must tend to zero on the claimed interval. Smoothness alone does not guarantee equality.
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Write the geometric foundation before substitution

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Construct a Maclaurin series for f(x)=1/(3+x²). Use the exact geometric foundation, not an already-substituted series. The rational function itself is defined for all real x.

(a) Write explicitly the Maclaurin identity for 1/(1−u), identify its first term and common ratio, and derive it from the finite sum through uᴺ. State its representation domain and test u=−1 and u=1.

LIM-8.F.1

(b) Use that identity to construct the complete infinite series for f, state its radius and interval, and test both endpoints. Evaluate the statement “f(2) exists, so its Maclaurin series must converge at 2.”

LIM-8.F.1

(c) At x=1/2, use the first three nonzero terms to approximate f. Give a guaranteed alternating error bound and compare it with the exact error using f(1/2)=4/13. Explain why the finite polynomial is not the infinite series.

LIM-8.F.1LIM-8.E.1

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Construct a new series from sine

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Let S(x)=sin(3x)/x for x≠0 and define S(0)=3. Angles are in radians. Use the sine Maclaurin foundation and ordinary series operations to construct a new infinite series. A supplied independent check is S(1/6)≈2.876553232.

(a) Write the sine foundation explicitly and use it to construct a complete indexed infinite Maclaurin series for S, not merely a finite polynomial. Give its first four nonzero terms, radius, representation interval, and explanation of the value at zero.

LIM-8.F.2

(b) Use the first three nonzero terms to estimate S(1/6). Verify the alternating magnitude condition, give a guaranteed error bound, and compare the bound with the observed difference from the supplied check. State whether the partial sum is above or below S(1/6).

LIM-8.F.2LIM-8.E.1

(c) Explain why multiplying the original sine-series coefficients by only 3 gives the wrong construction for sin(3x), and why dividing sin(3x) by x does not force a negative-power term here. Use the general term and the zero-value extension in your explanation.

LIM-8.F.2

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Construct an integral series from the exponential foundation

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Let W(x)=∫₀ˣ e^(−2t²)dt. Use the exponential Maclaurin foundation to represent W with a complete infinite series. For a numerical check at x=1/2, a calculator’s integral is approximately 0.427812196. The task does not require an elementary antiderivative.

The supplied graph of y=e^(−2t²) is positive and decreasing on 0≤t≤1. Shading from t=0 to 0.5 identifies the exact area W(0.5). The figure defines the integral but supplies no polynomial approximation or error bound.
The shaded region represents the exact integral W(1/2). Use the exponential foundation to construct a series approximation and a justified tail bound independently.
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The supplied graph of y=e^(−2t²) is positive and decreasing on 0≤t≤1. Shading from t=0 to 0.5 identifies the exact area W(0.5). The figure defines the integral but supplies no polynomial approximation or error bound.

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(a) Write the exponential foundation explicitly, substitute −2t², and integrate term by term to construct the complete infinite Maclaurin series for W. Give the first four nonzero terms, fix the constant using W(0), and justify its all-real representation interval.

LIM-8.F.2

(b) Estimate W(1/2) using the first three nonzero terms. Justify an alternating error bound, compare it with the supplied independent integral, and state whether the partial sum is an upper or lower estimate.

LIM-8.F.2LIM-8.E.1

(c) Differentiate your infinite series to check that it recovers e^(−2x²), and explain why W is increasing and odd. Use the supplied integrand graph to interpret W(1/2) as an area, and distinguish that area from the polynomial approximation in part b.

LIM-8.F.2

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For the teacher
  • Use derivative cycles to reconstruct the foundations. Ask for the representation interval whenever a series is written.

Unit planning and formative guidance →

Glossary and deeper reading · 19 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Polynomial

    A polynomial in one variable is a finite sum of constant multiples of nonnegative integer powers of that variable. The degree is the highest exponent with a nonzero coefficient. For example, 2x³−x+5 is a polynomial of degree 3. Expressions containing 1/x or √x are not polynomials in x.

  • Partial sum

    A partial sum adds the first finitely many terms of a series. The nth partial sum is Sₙ=a₁+a₂+⋯+aₙ when indexing begins at 1. For example, the first three terms of 1/2+1/4+1/8+⋯ give partial sum 7/8. Convergence of a series concerns the limit of these sums, not just the limit of the individual terms.

    Read more: 10.1 · Partial sum →

  • Derivative

    A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.

    Read more: 2.2 · Derivative →

  • Factorial

    The factorial n! multiplies the positive integers from 1 through n for a positive integer n. By definition, 0!=1. For example, 4!=4·3·2·1=24. Factorials occur in Taylor coefficients and their rapid growth often makes the ratio test useful.

  • Infinite series

    An infinite series is the sum of the terms of an infinite sequence, defined through its partial sums. The series converges when those partial sums approach a finite value. For example, 1/2+1/4+1/8+⋯ converges to 1. Terms approaching 0 are necessary for convergence, but that condition alone is insufficient.

    Read more: 10.1 · Infinite series →

  • Convergence of a series

    A series converges when the sequence of its partial sums approaches a finite limit. It diverges when there is no such finite limit. For example, 1/2+1/4+1/8+⋯ converges to 1, while the harmonic series diverges. Convergence of the individual terms to 0 is necessary but does not by itself establish convergence of their sum.

    Read more: 10.1 · Convergence of a series →

  • Geometric series

    A geometric series adds terms that have a fixed common ratio. With first term a and ratio r, it converges to a/(1−r) when |r|<1. For example, 1+1/2+1/4+⋯ sums to 2. For a nonzero first term and |r|≥1, the infinite series diverges.

    Read more: 10.2 · Geometric series →

  • Common ratio

    A common ratio is the fixed multiplier between consecutive terms of a geometric sequence or series. When the earlier term is nonzero, it equals the next term divided by that term. For example, 2,6,18,54 has common ratio 3. For a nonzero geometric series, its magnitude must be less than 1 for the series to converge.

  • Power series

    A power series is an infinite series of constant multiples of powers of x−c about a center c. It can converge for some inputs and diverge for others. For example, 1+x+x²+⋯ represents 1/(1−x) when |x|<1. Determining the radius and checking the endpoints establishes its interval of convergence.

    Read more: 10.13 · Power series →

  • Interval

    An interval is a set of real numbers containing every number between any two of its members. Parentheses exclude an endpoint, while square brackets include it. For example, (1,4] includes 4 and every number between 1 and 4, but excludes 1. An infinite endpoint always uses a parenthesis because infinity is not a real endpoint.

  • Radian

    A radian measures an angle by dividing the intercepted arc length by the circle’s radius. One full turn is 2π radians, so π radians equals 180 degrees. For example, an arc with length equal to its radius subtends an angle of one radian. The familiar derivative formulas for sine and cosine use radian inputs.

  • Limit

    A limit describes the value approached by outputs as inputs approach a specified target or grow without bound. For an input approaching a finite target, its value at that exact input does not determine the limit. For example, (x²−1)/(x−1) approaches 2 as x approaches 1 even though the original expression is undefined at 1. A finite two-sided limit requires the left and right approaches to agree.

    Read more: 1.2 · Limit →

  • Function

    A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

  • Integral

    Integration can mean accumulating a function over an interval or finding an antiderivative. A definite integral gives a number, while an indefinite integral represents a family of antiderivatives. For example, integrating 2x indefinitely gives x²+C, and integrating it from 0 to 2 gives 4. The Fundamental Theorem of Calculus connects these two operations under its hypotheses.

    Read more: 6.7 · Integral →

  • Integrand

    The integrand is the function or expression being integrated. The differential identifies the variable of integration. For example, in the integral of x² with respect to x, the integrand is x². Bounds belong to the integral’s interval rather than to the integrand itself.

  • Alternating series error bound

    For an alternating series satisfying the decreasing-to-zero magnitude conditions, the error after n terms is at most the next term’s magnitude. The actual sum lies between two consecutive partial sums. For example, stopping 1−1/2+1/3−⋯ after its tenth term gives error at most 1/11. The bound requires the alternating series hypotheses and does not apply automatically to every signed series.

    Read more: 10.10 · Alternating series error bound →

  • Rational function

    A rational function is a quotient of two polynomials. Its domain excludes inputs that make the original denominator zero. For example, (x²−1)/(x−1) agrees with x+1 away from x=1, but remains undefined there. Canceling a common factor does not automatically restore an excluded input.

  • Error bound

    An error bound gives a justified upper limit for the magnitude of an approximation’s error. It states how far the approximate value can be from the true value. For example, an error bound of 0.01 places the true value within 0.01 of the approximation. A bound requires its stated hypotheses and is not a claim that the error actually equals the bound.

  • Antiderivative

    An antiderivative of a function is another function whose derivative equals the original function on an interval. Two antiderivatives on the same interval differ by a constant. For example, x² is an antiderivative of 2x, and x²+7 is another. An antiderivative is a function, whereas a definite integral is an accumulated value.

    Read more: 6.8 · Antiderivative →

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