Chapter 8 Solutions

Authored worked explanation for the original source problem.

1. Replace infinity with B and integrate: ${{\int }_0}^B{e}^{-3x}dx=[-e^{-3x}/3{{]}_0}^B=(1-e^{-3B})/3$.

   Why: An infinite endpoint must be evaluated through a limit.

2. As B→∞, e^(−3B)→0, giving 1/3.

   Why: The positive decaying integrand has a finite accumulated area.

Reviewed answer: $1/3$

Takeaway: Infinity is a limiting endpoint, not a number to substitute.

Authored worked explanation for the original source problem.

1. Split $(x+3)/x^3=x^{-2}+3x^{-3}$.

   Why: Each term becomes a standard power.

2. Integrate to −x⁻¹−(3/2)x⁻²+C.

   Why: Increase each exponent by one and divide by the new exponent.

Reviewed answer: $-1/x-3/\left(2x^2\right)+C$, x≠0.

Takeaway: Split a polynomial numerator before choosing a technique.

Authored worked explanation for the original source problem.

Source review note: The source gives −3/x for the first term. The printed denominator is 3x⁴, so that term integrates to −1/(3x).

1. The integrand is $(1/3)x^{-2}+x^{-3}+x^{-4}$.

   Why: Divide every numerator term by 3x⁴.

2. Integrate each power to get the displayed primitive.

   Why: The coefficient 1/3 on the first term must remain.

3. Differentiating the answer reproduces the printed integrand.

   Why: This checks both coefficients and powers.

Reviewed answer: $-1/\left(3x\right)-1/\left(2x^2\right)-1/\left(3x^3\right)+C$, x≠0.

Takeaway: Distribute a denominator over every numerator term.

Authored worked explanation for the original source problem.

1. ${{\int }_0}^{B}3e^{-x/2}dx=[-6e^{-x/2}{{]}_0}^B=6(1-e^{-B/2})$.

   Why: Divide by the exponent slope −1/2 when finding the primitive.

2. Take B→∞ to get 6.

   Why: The exponential tail vanishes.

Reviewed answer: $6$

Takeaway: The decay coefficient affects the total area.

Authored worked explanation for the original source problem.

Source review note: The source gives +π/4. The printed integral includes a leading negative sign, so its reviewed value is −π/4.

1. A primitive is $-(1/2)\arcsin \left(2x\right)$ for −1/2<x<1/2.

   Why: Its derivative includes the chain factor 2 and the original negative sign.

2. At the singular endpoint use the right-hand limit of this primitive: $F\left(0\right)-{\lim }_{a\to -1/2^{+}}F\left(a\right)=0-\pi /4$.

   Why: Arcsine has a finite endpoint value even though the integrand is unbounded there.

Reviewed answer: $-\pi /4$

Takeaway: A negative integrand over increasing bounds gives a negative signed integral.

Authored worked explanation for the original source problem.

Source review note: The source gives (1/3)arctan(3x−2)+C, whose derivative is 1/((3x−2)²+1), lacking the printed square root. This source problem requires a logarithmic primitive beyond the listed inverse-trig patterns.

1. Complete the square: $9x^2-12x+5=(3x-2{)}^2+1$.

   Why: The radical is positive for all real x.

2. Let u=3x−2, giving one-third of $\int du/\sqrt{u^2+1}$, whose primitive is $\ln (u+\sqrt{u^2+1})$.

   Why: The log argument is positive because √(u²+1)>|u|.

3. Substitute back; differentiating the logarithm verifies the radical denominator.

   Why: The square root distinguishes this primitive from arctan's 1/(1+u²).

Reviewed answer: $(1/3)\ln (3x-2+\sqrt{(3x-2{)}^2+1})+C$

Takeaway: Recognize the exact denominator, including whether it has a square root.

Authored worked explanation for the original source problem.

Source review note: The source gives √5, but x²−5x−1 is negative on most of the printed integration interval. No real integral over [0,6] exists as written.

1. Require $x^2-5x-1>0$ for the square root in the denominator.

   Why: It must be positive rather than merely nonnegative because it divides the numerator.

2. Its roots are $(5\pm \sqrt{29})/2$; throughout [0,(5+√29)/2) the radicand is negative.

   Why: The positive root is about 5.19258, inside the stated interval.

3. The integrand is not real-defined on a whole portion of [0,6], so there is no real definite integral on that interval.

   Why: An endpoint limit cannot repair an entire interval of nonreal values.

Reviewed answer: The printed integral is not defined as a real integral over [0,6].

Takeaway: Check the integrand's domain across the full interval before integrating.

Authored worked explanation for the original source problem.

Source review note: The source primitive is consistent but omits +C in this indefinite integral.

1. Rewrite $x^4-8x^2+17=(x^2-4{)}^2+1$.

   Why: Complete the square in x².

2. The derivative of u=x²−4 is 2x, so the integrand is u′/(1+u²), the derivative of arctan u.

   Why: The numerator exactly matches the inner derivative.

Reviewed answer: $\arctan (x^2-4)+C$

Takeaway: An indefinite integral must include an arbitrary constant.

Authored worked explanation for the original source problem.

Source review note: The source gives arcsec(√(x²+1))+C, whose derivative is not the printed integrand. The reviewed logarithmic primitive differentiates correctly; this inherited problem uses a more advanced recognition than the chapter’s inverse-trig table.

1. Complete the square: $x^4+3x^2+2=(x^2+3/2{)}^2-1/4$.

   Why: Use x² as the intermediate variable.

2. With u=x²+3/2, du=2x dx; $\int du/\sqrt{u^2-1/4}=\ln (u+\sqrt{u^2-1/4})+C$.

   Why: u≥3/2, so the radical and logarithm are defined on the whole real line.

3. Substitute u back and verify by differentiation.

   Why: The derivative of the logarithmic primitive is exactly u′/√(u²−1/4).

Reviewed answer: $\ln (x^2+3/2+\sqrt{x^4+3x^2+2})+C$.

Takeaway: A radical of a completed square can require a log, not an inverse secant.

Authored worked explanation for the original source problem.

Source review note: The source expression is equivalent on 0<x<1 but writes √(1−x), so it does not extend to x>1, and omits +C. The reviewed form covers both original real intervals.

1. Let u=√x, so x=u² and dx=2u du. The integrand becomes 1/(u²−1) du.

   Why: The differential cancels the radical factor, while the original domain keeps x>0 and x≠1.

2. $1/(u^2-1)=(1/2)[1/(u-1)-1/(u+1\left)\right]$.

   Why: This algebraic identity can be checked by a common denominator.

3. Integrate to $(1/2)\ln \left|\right(u-1)/(u+1\left)\right|+C$ and substitute u=√x.

   Why: The absolute values give valid logs on either connected domain interval.

Reviewed answer: $(1/2)\ln \left|\right(\sqrt{x}-1)/(\sqrt{x}+1\left)\right|+C$ on 0<x<1 or x>1.

Takeaway: Keep the domain and constant while simplifying an equivalent primitive.

Authored solution to an added practice problem.

1. Integrate 3x², 4x, and −5 separately.

   Why: Linearity applies to an indefinite integral.

2. Their primitives are x³, 2x², and −5x, plus an arbitrary constant.

   Why: Differentiate the result to recover every term.

Reviewed answer: $x^3+2x^2-5x+C$

Takeaway: Check powers and coefficients by reversing the calculation.

Authored solution to an added practice problem.

1. A primitive of 2x is x².

   Why: Use the inverse power rule.

2. Evaluate $[x^2{{]}_1}^{\surd 10}=10-1=9$.

   Why: The upper bound's primitive value comes first.

Reviewed answer: $9$

Takeaway: Definite integration measures a difference of primitive values.

Authored solution to an added practice problem.

1. A primitive of 1/x is ln|x| on each sign interval.

   Why: The ordinary power formula does not apply at exponent −1.

2. Multiply by 3 and include C.

   Why: The constant multiple carries through integration.

Reviewed answer: $3\ln \left|x\right|+C$, x≠0.

Takeaway: Exponent −1 is the power-rule exception.

Authored solution to an added practice problem.

1. A primitive of sin x is −cos x and of e^x is e^x.

   Why: Use the familiar derivative pairs in reverse.

2. Keep the constant multiples 2 and 3 and add C.

   Why: Linearity combines the primitives.

Reviewed answer: $-2\cos x+3e^x+C$

Takeaway: Verify the cosine sign before finishing.

Authored solution to an added practice problem.

1. An antiderivative of x³ is x⁴/4.

   Why: Use the inverse power rule.

2. $[x^4/4{{]}_{-1}}^1=1/4-1/4=0$.

   Why: The negative and positive parts cancel as signed areas.

Reviewed answer: $0$

Takeaway: A zero signed integral need not mean zero geometric area.

Authored solution to an added practice problem.

1. The forward integral is ${{\int }_1}^{2}2xdx=4-1=3$.

   Why: Evaluate the easy forward direction first.

2. Reversing the bounds negates that value, giving −3.

   Why: The orientation of a definite integral matters.

Reviewed answer: $-3$

Takeaway: Reversed endpoints reverse the sign.

Authored solution to an added practice problem.

1. All primitives of 3x² are x³+C.

   Why: The integration constant is initially free.

2. F(2)=12 implies 8+C=12, so C=4.

   Why: The initial value picks the particular primitive.

Reviewed answer: $F\left(x\right)=x^3+4$.

Takeaway: Initial conditions are used after finding the primitive family.

Authored solution to an added practice problem.

1. ${{\int }_2}^B{x}^{-3}dx=[-1/(2x^2){{]}_2}^B=1/8-1/(2B^2)$.

   Why: The inverse power rule applies on x≥2.

2. Taking B→∞ gives 1/8.

   Why: The tail primitive vanishes at infinity.

Reviewed answer: $1/8$

Takeaway: A finite tail follows from a finite limiting primitive value.

Authored solution to an added practice problem.

1. ${{\int }_1}^{B}1/xdx=\ln B$.

   Why: The primitive is logarithmic, not a negative power.

2. ln B increases without bound as B→∞.

   Why: A decaying integrand alone does not guarantee a finite area.

Reviewed answer: Diverges to $+\infty$.

Takeaway: Decay must be fast enough for its primitive to approach a finite limit.

Authored solution to an added practice problem.

1. A primitive of e^(−x/2) is −2e^(−x/2).

   Why: The inside slope contributes the reciprocal factor.

2. At upper cutoff B the integral is $2e^{-1/2}-2e^{-B/2}$, which tends to 2e^(−1/2).

   Why: The offset lower endpoint retains its exponential value.

Reviewed answer: $2e^{-1/2}$

Takeaway: A tail starting later has less area.

Authored solution to an added practice problem.

1. Use a positive lower cutoff a: ${{\int }_a}^1{x}^{-1/2}dx=2-2\sqrt{a}$.

   Why: The integrand is unbounded at zero, so start with a finite interval inside its domain.

2. As a→0⁺, √a→0, giving 2.

   Why: The singular endpoint has a finite limiting area.

Reviewed answer: $2$

Takeaway: An infinite height can still have a finite improper area.

Authored solution to an added practice problem.

1. Write the denominator as 4[1+(x/2)²].

   Why: Match the standard inverse tangent pattern.

2. The proposed primitive differentiates to $(1/2)(1/(1+(x/2{)}^2)\left)\right(1/2)=1/(x^2+4)$.

   Why: Both scale factors are needed.

Reviewed answer: $(1/2)\arctan (x/2)+C$

Takeaway: Use a derivative check to keep scale factors correct.

Authored solution to an added practice problem.

1. Factor the radical as $3\sqrt{1-(x/3{)}^2}$.

   Why: The positive factor 3 comes outside the square root.

2. The derivative of arcsin(x/3) is $(1/3)/\sqrt{1-(x/3{)}^2}$, exactly the integrand.

   Why: The inner 1/3 cancels the radical's scale.

Reviewed answer: $\arcsin (x/3)+C$, for |x|<3.

Takeaway: A scaled arcsine can absorb the full radical coefficient.

Authored solution to an added practice problem.

1. Complete the square: x²−4x+5=(x−2)²+1.

   Why: A shift reveals the standard pattern.

2. The inner x−2 has derivative 1, so no extra factor is needed.

   Why: Differentiate arctan(x−2) to check.

Reviewed answer: $\arctan (x-2)+C$

Takeaway: Complete the square before searching for an inverse-trig primitive.

Authored solution to an added practice problem.

1. A primitive is arctan(x²), since its derivative is 2x/(1+x⁴).

   Why: The numerator matches the inside derivative.

2. $\left[\arctan \right(x^2){{]}_0}^1=\pi /4-0$.

   Why: The definite endpoints have familiar inverse tangent values.

Reviewed answer: $\pi /4$

Takeaway: Spot an inside derivative multiplying a familiar outer pattern.

Authored solution to an added practice problem.

1. For θ in [0,π/6], dx=cosθ dθ and √(1−x²)=cosθ.

   Why: The chosen angle interval keeps cosine nonnegative.

2. At x=0, θ=0; at x=1/2, θ=π/6. Substitution cancels the cosine factor.

   Why: Change the variable, differential, and both bounds together.

Reviewed answer: With x=sinθ, the integral becomes ${{\int }_0}^{\pi /6}\sin \theta d\theta$.

Takeaway: A definite substitution has three coordinated changes.

Authored solution to an added practice problem.

1. The integrand √(1+t²) is continuous for every real t.

   Why: The Fundamental Theorem of Calculus applies on any finite interval.

2. An accumulation from a fixed bound to x differentiates to the integrand at x.

   Why: The moving upper bound supplies the current endpoint height.

Reviewed answer: $F\prime \left(x\right)=\sqrt{1+x^2}$.

Takeaway: An accumulation derivative can be easier than finding a primitive.

Authored solution to an added practice problem.

1. Let G(u)=∫₀ᵘcos t dt, so G′(u)=cos u.

   Why: The accumulation's derivative is the endpoint integrand.

2. F(x)=G(x²), giving F′=G′(x²)·2x.

   Why: The moving endpoint itself requires the chain rule.

Reviewed answer: $F\prime \left(x\right)=2x\cos \left(x^2\right)$.

Takeaway: A nontrivial upper endpoint contributes its derivative.

Authored solution to an added practice problem.

1. Every x²+C differentiates to 2x.

   Why: Differentiation erases additive constants.

2. The primitive x² alone omits the other family members; the given initial value sets C=3.

   Why: An indefinite integral and a selected primitive are different outputs.

Reviewed answer: The complete family is $x^2+C$; with F(0)=3 it becomes x²+3.

Takeaway: Write the family before applying extra information.

Authored solution to an added practice problem.

1. A primitive of 1/√(1−x²) is arcsin x on (−1,1).

   Why: The integration interval lies inside its finite-derivative domain.

2. $\arcsin (1/2)-\arcsin \left(0\right)=\pi /6$.

   Why: Use principal angles in radians.

Reviewed answer: $\pi /6$

Takeaway: The principal inverse value fixes the exact definite result.