Calculus 1 Quiz: U Substitution
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U SubstitutionQuestion 1 of 20

Let f(x)f(x) be an antiderivative of g(x)=5xe−x2g(x) = 5x e^{-x^2}. If f(0)=3f(0) = 3, what is the value of f(1)f(1)?

−52e−1-\frac{5}{2}e^{-1}
1+5e−12\frac{1 + 5e^{-1}}{2}
8−5e−18 - 5e^{-1}
11−5e−12\frac{11 - 5e^{-1}}{2}
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Calculus 1 Quiz

Calculus 1 Quiz: U Substitution

Practice U Substitution in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on U Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let f(x)f(x) be an antiderivative of g(x)=5xe−x2g(x) = 5x e^{-x^2}. If f(0)=3f(0) = 3, what is the value of f(1)f(1)?

  1. −52e−1-\frac{5}{2}e^{-1}
  2. 1+5e−12\frac{1 + 5e^{-1}}{2}
  3. 8−5e−18 - 5e^{-1}
  4. 11−5e−12\frac{11 - 5e^{-1}}{2} (correct answer)
Explanation: When you see a problem asking for the value of an antiderivative at a specific point, you need to use the Fundamental Theorem of Calculus along with the given initial condition. Since f(x)f(x) is an antiderivative of g(x)=5xe−x2g(x) = 5xe^{-x^2}, you know that f′(x)=g(x)f'(x) = g(x). To find f(1)f(1), you can use: f(1)=f(0)+∫01g(x) dx=3+∫015xe−x2 dxf(1) = f(0) + \int_0^1 g(x)\,dx = 3 + \int_0^1 5xe^{-x^2}\,dx. For the integral ∫015xe−x2 dx\int_0^1 5xe^{-x^2}\,dx, use substitution with u=−x2u = -x^2, so du=−2x dxdu = -2x\,dx, which means x dx=−12dux\,dx = -\frac{1}{2}du. When x=0x = 0, u=0u = 0; when x=1x = 1, u=−1u = -1. The integral becomes: 5∫0−1eu⋅(−12)du=−52∫0−1eu du=−52[eu]0−1=−52(e−1−1)=−52e−1+525\int_0^{-1} e^u \cdot \left(-\frac{1}{2}\right)du = -\frac{5}{2}\int_0^{-1} e^u\,du = -\frac{5}{2}[e^u]_0^{-1} = -\frac{5}{2}(e^{-1} - 1) = -\frac{5}{2}e^{-1} + \frac{5}{2}. Therefore: f(1)=3+52−52e−1=112−5e−12=11−5e−12f(1) = 3 + \frac{5}{2} - \frac{5}{2}e^{-1} = \frac{11}{2} - \frac{5e^{-1}}{2} = \frac{11 - 5e^{-1}}{2}. Answer D is correct. Answer A gives only the integral value without adding f(0)f(0). Answer B incorrectly handles the substitution bounds. Answer C likely comes from computational errors in the integration process. Study tip: When finding antiderivative values, always remember that F(b)=F(a)+∫abF′(x) dxF(b) = F(a) + \int_a^b F'(x)\,dx. The initial condition is crucial—don't forget to add it to your integral result.

Question 2

Evaluate ∫01x+1x2+1 dx\int_0^1 \frac{x+1}{x^2+1} \, dx.

  1. 12ln⁡(2)+π4\frac{1}{2}\ln(2) + \frac{\pi}{4} (correct answer)
  2. ln⁡(2)+π4\ln(2) + \frac{\pi}{4}
  3. 12ln⁡(2)\frac{1}{2}\ln(2)
  4. π4\frac{\pi}{4}
Explanation: First, split the integral into two parts: ∫01xx2+1 dx+∫011x2+1 dx\int_0^1 \frac{x}{x^2+1} \, dx + \int_0^1 \frac{1}{x^2+1} \, dx. For the first integral, use u-substitution with u=x2+1u=x^2+1, du=2x dxdu=2x\,dx. This gives 12∫121u du=12[ln⁡∣u∣]12=12(ln⁡(2)−ln⁡(1))=12ln⁡(2)\frac{1}{2}\int_1^2 \frac{1}{u} \, du = \frac{1}{2}[\ln|u|]_1^2 = \frac{1}{2}(\ln(2)-\ln(1)) = \frac{1}{2}\ln(2). The second integral is a standard form: [arctan⁡(x)]01=arctan⁡(1)−arctan⁡(0)=π4−0=π4[\arctan(x)]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. The final answer is the sum of the two results: 12ln⁡(2)+π4\frac{1}{2}\ln(2) + \frac{\pi}{4}.

Question 3

Which of the following integrals is equivalent to ∫0π/3tan⁡(x) dx\int_0^{\pi/3} \tan(x) \, dx?

  1. ∫11/21u du\int_1^{1/2} \frac{1}{u} \, du
  2. ∫03/21u du\int_0^{\sqrt{3}/2} \frac{1}{u} \, du
  3. ∫11/2−1u du\int_1^{1/2} -\frac{1}{u} \, du (correct answer)
  4. ∫0π/3u du\int_0^{\pi/3} u \, du
Explanation: When you encounter an integral that looks difficult to evaluate directly, substitution is often the key technique to simplify it. For ∫0π/3tan⁡(x) dx\int_0^{\pi/3} \tan(x) \, dx, you need to use the fact that tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)} and make a strategic substitution. Let u=cos⁡(x)u = \cos(x), so du=−sin⁡(x) dxdu = -\sin(x) \, dx, which means sin⁡(x) dx=−du\sin(x) \, dx = -du. Now you can rewrite the integral as: ∫0π/3sin⁡(x)cos⁡(x) dx=∫1u⋅(−du)=∫−1u du\int_0^{\pi/3} \frac{\sin(x)}{\cos(x)} \, dx = \int \frac{1}{u} \cdot (-du) = \int -\frac{1}{u} \, du The crucial step is updating the limits of integration. When x=0,u=cos⁡(0)=1x = 0,u = \cos(0) = 1. When x=π/3,u=cos⁡(π/3)=1/2x = \pi/3,u = \cos(\pi/3) = 1/2. Therefore, the equivalent integral is ∫11/2−1u du\int_1^{1/2} -\frac{1}{u} \, du. Option A, ∫11/21u du\int_1^{1/2} \frac{1}{u} \, du, has the correct limits but is missing the negative sign from the substitution. Option B, ∫03/21u du\int_0^{\sqrt{3}/2} \frac{1}{u} \, du, uses incorrect limits—the lower limit should be 1, not 0, and cos⁡(π/3)=1/2\cos(\pi/3) = 1/2, not 3/2\sqrt{3}/2. Option D completely abandons the substitution approach and doesn't represent the same function. Study tip: When using substitution, always remember to change both the integrand AND the limits of integration. The negative sign from du=−sin⁡(x) dxdu = -\sin(x) \, dx is easy to forget but crucial for getting the right answer.

Question 4

To evaluate the integral ∫x+1x2+2x+5 dx\int \frac{x+1}{x^2+2x+5} \, dx, the most effective first step is the substitution u=x2+2x+5u = x^2+2x+5. Which of the following is the resulting integral in terms of uu?

  1. ∫1u du\int \frac{1}{u} \, du
  2. ∫2u du\int \frac{2}{u} \, du
  3. ∫12u du\int \frac{1}{2u} \, du (correct answer)
  4. ∫u−4u du\int \frac{u-4}{u} \, du
Explanation: Let u=x2+2x+5u = x^2+2x+5. Then the derivative is dudx=2x+2=2(x+1)\frac{du}{dx} = 2x+2 = 2(x+1). Rearranging gives du=2(x+1) dxdu = 2(x+1)\,dx, or (x+1) dx=12du(x+1)\,dx = \frac{1}{2}du. Substituting both uu and 12du\frac{1}{2}du into the original integral gives ∫1u⋅12du=∫12u du\int \frac{1}{u} \cdot \frac{1}{2}du = \int \frac{1}{2u} \, du.

Question 5

Find the indefinite integral ∫xx+3 dx\int x \sqrt{x+3} \, dx.

  1. 25(x+3)5/2−2(x+3)3/2+C\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C (correct answer)
  2. 23x(x+3)3/2+C\frac{2}{3}x(x+3)^{3/2} + C
  3. 12x2⋅23(x+3)3/2+C\frac{1}{2}x^2 \cdot \frac{2}{3}(x+3)^{3/2} + C
  4. 25(x+3)5/2+2(x+3)3/2+C\frac{2}{5}(x+3)^{5/2} + 2(x+3)^{3/2} + C
Explanation: Let u=x+3u = x+3. Then du=dxdu = dx and x=u−3x = u-3. The integral becomes ∫(u−3)u du=∫(u3/2−3u1/2) du\int (u-3)\sqrt{u} \, du = \int (u^{3/2} - 3u^{1/2}) \, du. Integrating with respect to uu gives 25u5/2−3(23u3/2)+C=25u5/2−2u3/2+C\frac{2}{5}u^{5/2} - 3\left(\frac{2}{3}u^{3/2}\right) + C = \frac{2}{5}u^{5/2} - 2u^{3/2} + C. Substituting back u=x+3u=x+3 yields 25(x+3)5/2−2(x+3)3/2+C\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C.

Question 6

Find the indefinite integral ∫(x+1x)3(1−1x2) dx\int (x + \frac{1}{x})^3 (1 - \frac{1}{x^2}) \, dx.

  1. 14(1−1x2)4+C\frac{1}{4}(1 - \frac{1}{x^2})^4 + C
  2. (x+1x)4+C(x + \frac{1}{x})^4 + C
  3. 3(x+1x)2+C3(x + \frac{1}{x})^2 + C
  4. 14(x+1x)4+C\frac{1}{4}(x + \frac{1}{x})^4 + C (correct answer)
Explanation: When you encounter an integral with a complicated expression raised to a power multiplied by another expression, look for substitution opportunities. The key insight here is recognizing that the derivative of (x+1x)(x + \frac{1}{x}) appears elsewhere in the integrand. Let's use substitution with u=x+1xu = x + \frac{1}{x}. Taking the derivative: dudx=1−1x2\frac{du}{dx} = 1 - \frac{1}{x^2}, which means du=(1−1x2)dxdu = (1 - \frac{1}{x^2})dx. This is exactly the second factor in our integrand! Substituting into the original integral: ∫(x+1x)3(1−1x2) dx=∫u3 du\int (x + \frac{1}{x})^3 (1 - \frac{1}{x^2}) \, dx = \int u^3 \, du This simplifies to: ∫u3 du=u44+C=14(x+1x)4+C\int u^3 \, du = \frac{u^4}{4} + C = \frac{1}{4}(x + \frac{1}{x})^4 + C Answer choice A, 14(1−1x2)4+C\frac{1}{4}(1 - \frac{1}{x^2})^4 + C, incorrectly treats (1−1x2)(1 - \frac{1}{x^2}) as the substitution variable instead of recognizing it as the derivative. Answer choice B, (x+1x)4+C(x + \frac{1}{x})^4 + C, forgets the coefficient 14\frac{1}{4} that comes from integrating u3u^3. Answer choice C, 3(x+1x)2+C3(x + \frac{1}{x})^2 + C, appears to confuse integration with differentiation, giving something closer to a derivative form. Study tip: When you see a composite function raised to a power multiplied by another expression, immediately check if that other expression might be the derivative of the inner function. This pattern appears frequently in calculus and makes substitution problems much more manageable.

Question 7

Find the antiderivative of f(x)=sin⁡(x)xf(x) = \frac{\sin(\sqrt{x})}{\sqrt{x}}.

  1. −12cos⁡(x)+C-\frac{1}{2}\cos(\sqrt{x}) + C
  2. 12cos⁡(x)+C\frac{1}{2}\cos(\sqrt{x}) + C
  3. 2cos⁡(x)+C2\cos(\sqrt{x}) + C
  4. −2cos⁡(x)+C-2\cos(\sqrt{x}) + C (correct answer)
Explanation: When you encounter an antiderivative involving a composite function like sin⁡(x)x\frac{\sin(\sqrt{x})}{\sqrt{x}}, substitution is your go-to technique. The key insight is recognizing that the denominator x\sqrt{x} is related to the derivative of the inner function x\sqrt{x} in the sine term. Let's use substitution with u=xu = \sqrt{x}. Then dudx=12x\frac{du}{dx} = \frac{1}{2\sqrt{x}}, which means dx=2x du=2u dudx = 2\sqrt{x} \, du = 2u \, du. Substituting into our integral: ∫sin⁡(x)x dx=∫sin⁡(u)u⋅2u du=∫2sin⁡(u) du=−2cos⁡(u)+C\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{\sin(u)}{u} \cdot 2u \, du = \int 2\sin(u) \, du = -2\cos(u) + C Converting back to xx: −2cos⁡(x)+C-2\cos(\sqrt{x}) + C. Now let's examine why the other answers are incorrect. Choice A gives −12cos⁡(x)+C-\frac{1}{2}\cos(\sqrt{x}) + C, which results from incorrectly handling the substitution factor of 2. Choice B, 12cos⁡(x)+C\frac{1}{2}\cos(\sqrt{x}) + C, makes the same substitution error plus incorrectly integrates sine (forgetting the negative sign). Choice C, 2cos⁡(x)+C2\cos(\sqrt{x}) + C, correctly handles the factor of 2 but misses the negative sign that comes from integrating sine. The correct answer is D: −2cos⁡(x)+C-2\cos(\sqrt{x}) + C. Study tip: When dealing with composite functions in integration, always check if u-substitution applies. Look for the derivative of the inner function appearing elsewhere in the integrand—it's often disguised but present, as 1x\frac{1}{\sqrt{x}} was here.

Question 8

Using the substitution u=tan⁡(x)u = \tan(x), the integral ∫0π/4tan⁡3(x)sec⁡2(x)dx\int_0^{\pi/4} \tan^3(x) \sec^2(x) dx is transformed into which of the following definite integrals?

  1. ∫12(u3−u)du\int_1^{\sqrt{2}} (u^3 - u) du
  2. ∫0π/4u3du\int_0^{\pi/4} u^3 du
  3. ∫01u31+u2du\int_0^1 \frac{u^3}{1+u^2} du
  4. ∫01u3du\int_0^1 u^3 du (correct answer)
Explanation: When you encounter integrals involving powers of trigonometric functions with their derivatives present, substitution is often the key strategy. Here, the presence of sec⁡2(x)\sec^2(x) alongside powers of tan⁡(x)\tan(x) strongly suggests using u=tan⁡(x)u = \tan(x) since dudx=sec⁡2(x)\frac{du}{dx} = \sec^2(x). Let's work through the substitution systematically. With u=tan⁡(x)u = \tan(x), we have du=sec⁡2(x)dxdu = \sec^2(x) dx. The integral ∫0π/4tan⁡3(x)sec⁡2(x)dx\int_0^{\pi/4} \tan^3(x) \sec^2(x) dx becomes ∫tan⁡3(x)sec⁡2(x)dx=∫u3du\int \tan^3(x) \sec^2(x) dx = \int u^3 du. For the limits of integration: when x=0,u=tan⁡(0)=0x = 0,u = \tan(0) = 0; when x=π/4,u=tan⁡(π/4)=1x = \pi/4,u = \tan(\pi/4) = 1. Therefore, the transformed integral is ∫01u3du\int_0^1 u^3 du, which is answer D. Let's examine why the other options are incorrect. Choice A, ∫12(u3−u)du\int_1^{\sqrt{2}} (u^3 - u) du, uses wrong limits (2\sqrt{2} would come from sec⁡(π/4)\sec(\pi/4), not tan⁡(π/4)\tan(\pi/4)) and introduces an extra −u-u term. Choice B, ∫0π/4u3du\int_0^{\pi/4} u^3 du, fails to transform the limits of integration—they should change from xx-values to uu-values. Choice C, ∫01u31+u2du\int_0^1 \frac{u^3}{1+u^2} du, has correct limits but unnecessarily complicates the integrand with 1+u21+u^2 in the denominator. Remember: when using substitution, always transform both the integrand AND the limits of integration. The derivative of your substitution variable should appear somewhere in the original integral.

Question 9

What is the average value of the function f(x)=x(2x2+1)2f(x) = \frac{x}{(2x^2+1)^2} on the interval [0,1][0, 1]?

  1. 23\frac{2}{3}
  2. 16\frac{1}{6} (correct answer)
  3. −13-\frac{1}{3}
  4. ln⁡(3)2\frac{\ln(3)}{2}
Explanation: When you see "average value of a function," you need to use the formula: Average=1b−a∫abf(x) dx\text{Average} = \frac{1}{b-a}\int_a^b f(x)\,dx. Here, you're finding the average of f(x)=x(2x2+1)2f(x) = \frac{x}{(2x^2+1)^2} on [0,1][0,1]. First, evaluate ∫01x(2x2+1)2 dx\int_0^1 \frac{x}{(2x^2+1)^2}\,dx. Use substitution: let u=2x2+1u = 2x^2 + 1, so du=4x dxdu = 4x\,dx, which means x dx=14dux\,dx = \frac{1}{4}du. When x=0,u=1x = 0,u = 1; when x=1,u=3x = 1,u = 3. The integral becomes: ∫131u2⋅14 du=14∫13u−2 du=14[−u−1]13=14(−13+1)=14⋅23=16\int_1^3 \frac{1}{u^2} \cdot \frac{1}{4}\,du = \frac{1}{4}\int_1^3 u^{-2}\,du = \frac{1}{4}[-u^{-1}]_1^3 = \frac{1}{4}\left(-\frac{1}{3} + 1\right) = \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{6} Since the interval length is 1−0=11-0 = 1, the average value is 11⋅16=16\frac{1}{1} \cdot \frac{1}{6} = \frac{1}{6}, confirming answer B. Answer A (23\frac{2}{3}) likely comes from forgetting the 14\frac{1}{4} factor in the substitution. Answer C (−13-\frac{1}{3}) represents a sign error during integration. Answer D (ln⁡(3)2\frac{\ln(3)}{2}) suggests incorrectly treating this as ∫1u du=ln⁡∣u∣\int \frac{1}{u}\,du = \ln|u| instead of ∫u−2 du\int u^{-2}\,du. Study tip: For average value problems, always double-check your u-substitution coefficients and remember that the average value formula requires dividing by the interval length.

Question 10

If u=ex+1u = e^x + 1, which of the following integrals is equivalent to ∫01exex+1 dx\int_0^1 \frac{e^x}{e^x+1} \, dx?

  1. ∫011u du\int_0^1 \frac{1}{u} \, du
  2. ∫121u du\int_1^2 \frac{1}{u} \, du
  3. ∫2e+11u du\int_2^{e+1} \frac{1}{u} \, du (correct answer)
  4. ∫1e1u du\int_1^{e} \frac{1}{u} \, du
Explanation: Given the substitution u=ex+1u = e^x + 1, we find the differential du=exdxdu = e^x dx. Now, we must transform the limits of integration. The lower limit is x=0x=0, so u=e0+1=1+1=2u = e^0 + 1 = 1+1=2. The upper limit is x=1x=1, so u=e1+1=e+1u = e^1 + 1 = e+1. Substituting uu and dudu into the integral, we get ∫2e+11u du\int_2^{e+1} \frac{1}{u} \, du.

Question 11

Let ff be a continuous function such that ∫09f(x)dx=4\int_0^9 f(x) dx = 4. What is the value of ∫03xf(x2)dx\int_0^3 x f(x^2) dx?

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 8
Explanation: To evaluate the integral ∫03xf(x2)dx\int_0^3 x f(x^2) dx, we use a u-substitution. Let u=x2u = x^2. Then, the differential is du=2xdxdu = 2x dx, which implies xdx=12dux dx = \frac{1}{2} du. We must also change the limits of integration. The lower limit is x=0x=0, so u=02=0u = 0^2 = 0. The upper limit is x=3x=3, so u=32=9u = 3^2 = 9. Substituting these into the integral gives: ∫03xf(x2)dx=∫09f(u)(12du)=12∫09f(u)du\int_0^3 x f(x^2) dx = \int_0^9 f(u) \left(\frac{1}{2} du\right) = \frac{1}{2} \int_0^9 f(u) du Since the variable of integration is a dummy variable, ∫09f(u)du\int_0^9 f(u) du is equivalent to ∫09f(x)dx\int_0^9 f(x) dx. We are given that ∫09f(x)dx=4\int_0^9 f(x) dx = 4. Therefore, the value of the integral is 12(4)=2\frac{1}{2} (4) = 2.

Question 12

Which of the following is an antiderivative of f(x)=e3x1+e6xf(x) = \frac{e^{3x}}{1+e^{6x}}?

  1. 16ln⁡(1+e6x)+C\frac{1}{6}\ln(1+e^{6x}) + C
  2. 13arctan⁡(e3x)+C\frac{1}{3}\arctan(e^{3x}) + C (correct answer)
  3. 3arctan⁡(e3x)+C3\arctan(e^{3x}) + C
  4. 13arcsin⁡(e3x)+C\frac{1}{3}\arcsin(e^{3x}) + C
Explanation: We need to evaluate ∫e3x1+e6x dx\int \frac{e^{3x}}{1+e^{6x}} \, dx. Notice that e6x=(e3x)2e^{6x} = (e^{3x})^2. Let u=e3xu = e^{3x}. Then du=3e3x dxdu = 3e^{3x} \, dx, so e3x dx=13due^{3x} \, dx = \frac{1}{3}du. The integral becomes ∫11+u2⋅13du=13∫11+u2 du=13arctan⁡(u)+C\int \frac{1}{1+u^2} \cdot \frac{1}{3}du = \frac{1}{3} \int \frac{1}{1+u^2} \, du = \frac{1}{3}\arctan(u) + C. Substituting back u=e3xu=e^{3x} gives 13arctan⁡(e3x)+C\frac{1}{3}\arctan(e^{3x}) + C.

Question 13

Evaluate ∫cos⁡(x)sin⁡2(x) dx\int \frac{\cos(x)}{\sin^2(x)} \, dx.

  1. ln⁡(sin⁡2(x))+C\ln(\sin^2(x)) + C
  2. −csc⁡(x)+C-\csc(x) + C (correct answer)
  3. cot⁡(x)+C\cot(x) + C
  4. −13sin⁡3(x)+C-\frac{1}{3\sin^3(x)} + C
Explanation: When you encounter an integral with trigonometric functions where one appears in the numerator and a power of another in the denominator, look for substitution opportunities or recognize standard antiderivative patterns. For ∫cos⁡(x)sin⁡2(x) dx\int \frac{\cos(x)}{\sin^2(x)} \, dx, notice that the derivative of sin⁡(x)\sin(x) is cos⁡(x)\cos(x), which appears in the numerator. This suggests using substitution with u=sin⁡(x)u = \sin(x), so du=cos⁡(x) dxdu = \cos(x) \, dx. The integral becomes ∫1u2 du=∫u−2 du=−u−1+C=−1sin⁡(x)+C=−csc⁡(x)+C\int \frac{1}{u^2} \, du = \int u^{-2} \, du = -u^{-1} + C = -\frac{1}{\sin(x)} + C = -\csc(x) + C. Alternatively, you can recognize this directly as the derivative pattern: since ddx[−csc⁡(x)]=csc⁡(x)cot⁡(x)=1sin⁡(x)⋅cos⁡(x)sin⁡(x)=cos⁡(x)sin⁡2(x)\frac{d}{dx}[-\csc(x)] = \csc(x)\cot(x) = \frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{\cos(x)}{\sin^2(x)}, the antiderivative is −csc⁡(x)+C-\csc(x) + C. Choice A, ln⁡(sin⁡2(x))+C\ln(\sin^2(x)) + C, would have derivative 2cos⁡(x)sin⁡(x)\frac{2\cos(x)}{\sin(x)}, not our integrand. Choice C, cot⁡(x)+C\cot(x) + C, has derivative −csc⁡2(x)=−1sin⁡2(x)-\csc^2(x) = -\frac{1}{\sin^2(x)}, missing the cos⁡(x)\cos(x) factor. Choice D, −13sin⁡3(x)+C-\frac{1}{3\sin^3(x)} + C, would come from integrating cos⁡(x)sin⁡4(x)\frac{\cos(x)}{\sin^4(x)}, not sin⁡2(x)\sin^2(x). Study tip: When you see cos⁡(x)\cos(x) in the numerator with powers of sin⁡(x)\sin(x) in the denominator, immediately consider uu-substitution with u=sin⁡(x)u = \sin(x). Also memorize that ddx[csc⁡(x)]=−csc⁡(x)cot⁡(x)\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x).

Question 14

The rate of a chemical reaction, in moles per second, is given by R(t)=tet2R(t) = \frac{t}{e^{t^2}}, where tt is time in seconds. What is the total number of moles produced during the first second (from t=0t=0 to t=1t=1)?

  1. 12(1−1e)\frac{1}{2}(1 - \frac{1}{e}) (correct answer)
  2. 1−1e1 - \frac{1}{e}
  3. 1e\frac{1}{e}
  4. 12(1e−1)\frac{1}{2}(\frac{1}{e} - 1)
Explanation: The total moles produced is the integral of the rate: ∫01te−t2 dt\int_0^1 t e^{-t^2} \, dt. Let u=−t2u = -t^2. Then du=−2t dtdu = -2t \, dt, so t dt=−12dut \, dt = -\frac{1}{2}du. Change the limits: when t=0t=0, u=0u=0. When t=1t=1, u=−1u=-1. The integral becomes −12∫0−1eu du=12∫−10eu du=12[eu]−10=12(e0−e−1)=12(1−1e)-\frac{1}{2} \int_0^{-1} e^u \, du = \frac{1}{2} \int_{-1}^0 e^u \, du = \frac{1}{2}[e^u]_{-1}^0 = \frac{1}{2}(e^0 - e^{-1}) = \frac{1}{2}(1 - \frac{1}{e}).

Question 15

Evaluate ∫0π/6sin⁡(x)cos⁡3(x) dx\int_0^{\pi/6} \sin(x) \cos^3(x) \, dx.

  1. 7/647/64 (correct answer)
  2. −7/64-7/64
  3. 1/41/4
  4. 15/6415/64
Explanation: Let u=cos⁡(x)u = \cos(x). Then du=−sin⁡(x) dxdu = -\sin(x) \, dx, so sin⁡(x) dx=−du\sin(x) \, dx = -du. When x=0x=0, u=cos⁡(0)=1u=\cos(0)=1. When x=π/6x=\pi/6, u=cos⁡(π/6)=3/2u=\cos(\pi/6)=\sqrt{3}/2. The integral transforms to ∫13/2u3(−du)=−∫13/2u3 du=∫3/21u3 du\int_1^{\sqrt{3}/2} u^3 (-du) = -\int_1^{\sqrt{3}/2} u^3 \, du = \int_{\sqrt{3}/2}^1 u^3 \, du. Evaluating this gives [u44]3/21=144−(3/2)44=14−9/164=14−964=1664−964=764\left[\frac{u^4}{4}\right]_{\sqrt{3}/2}^1 = \frac{1^4}{4} - \frac{(\sqrt{3}/2)^4}{4} = \frac{1}{4} - \frac{9/16}{4} = \frac{1}{4} - \frac{9}{64} = \frac{16}{64} - \frac{9}{64} = \frac{7}{64}.

Question 16

Which of the following is the indefinite integral of f(x)=sin⁡(x)cos⁡(cos⁡(x))f(x) = \sin(x) \cos(\cos(x))?

  1. −sin⁡(cos⁡(x))+C-\sin(\cos(x)) + C (correct answer)
  2. sin⁡(cos⁡(x))+C\sin(\cos(x)) + C
  3. cos⁡(cos⁡(x))+C\cos(\cos(x)) + C
  4. 12sin⁡2(x)+C\frac{1}{2}\sin^2(x) + C
Explanation: To find the integral ∫sin⁡(x)cos⁡(cos⁡(x))dx\int \sin(x) \cos(\cos(x)) dx, we use u-substitution. Let the inner function be u=cos⁡(x)u = \cos(x). Then, its differential is du=−sin⁡(x)dxdu = -\sin(x) dx. This means sin⁡(x)dx=−du\sin(x) dx = -du. Substituting these into the integral gives: ∫cos⁡(cos⁡(x))(sin⁡(x)dx)=∫cos⁡(u)(−du)=−∫cos⁡(u)du\int \cos(\cos(x)) (\sin(x) dx) = \int \cos(u) (-du) = -\int \cos(u) du The integral of cos⁡(u)\cos(u) is sin⁡(u)\sin(u). Therefore: −sin⁡(u)+C-\sin(u) + C Substituting back for u=cos⁡(x)u = \cos(x), we get the final answer: −sin⁡(cos⁡(x))+C-\sin(\cos(x)) + C

Question 17

Evaluate the definite integral ∫02x(x2+1)2dx\int_0^2 \frac{x}{(x^2+1)^2} dx

  1. 25\frac{2}{5} (correct answer)
  2. −25-\frac{2}{5}
  3. 45\frac{4}{5}
  4. 12\frac{1}{2}
Explanation: Let u=x2+1u = x^2+1. Then du=2x dxdu = 2x \, dx, which implies x dx=12dux \, dx = \frac{1}{2} du. We must also change the limits of integration. When x=0x=0, u=02+1=1u = 0^2+1 = 1. When x=2x=2, u=22+1=5u = 2^2+1 = 5. The integral transforms to: 12∫151u2du=12∫15u−2du=12[−u−1]15=12[−1u]15\frac{1}{2} \int_1^5 \frac{1}{u^2} du = \frac{1}{2} \int_1^5 u^{-2} du = \frac{1}{2} \left[ -u^{-1} \right]_1^5 = \frac{1}{2} \left[ -\frac{1}{u} \right]_1^5 Evaluating this gives: 12(−15−(−11))=12(−15+1)=12(45)=25\frac{1}{2} \left( -\frac{1}{5} - \left(-\frac{1}{1}\right) \right) = \frac{1}{2} \left( -\frac{1}{5} + 1 \right) = \frac{1}{2} \left( \frac{4}{5} \right) = \frac{2}{5}

Question 18

For which of the following integrals is the substitution u=x2+4u = x^2 + 4 the most direct and effective first step for finding the antiderivative?

  1. ∫x3x2+4dx\int \frac{x^3}{x^2 + 4} dx
  2. ∫1x2+4dx\int \frac{1}{x^2 + 4} dx
  3. ∫x(x2+4)2dx\int \frac{x}{(x^2 + 4)^2} dx (correct answer)
  4. ∫x2x2+4dx\int \frac{x^2}{\sqrt{x^2 + 4}} dx
Explanation: The substitution u=g(x)u = g(x) is most direct when the integrand is of the form f(g(x))g′(x)f(g(x))g'(x). Here, g(x)=x2+4g(x) = x^2+4, so g′(x)=2xg'(x) = 2x. We are looking for an integral with a factor of xx in the numerator and the rest of the expression being a function of x2+4x^2+4. For choice C, ∫x(x2+4)2dx\int \frac{x}{(x^2 + 4)^2} dx, if we let u=x2+4u = x^2+4, then du=2x dxdu = 2x \, dx, so x dx=12dux \, dx = \frac{1}{2} du. The integral becomes 12∫1u2du\frac{1}{2} \int \frac{1}{u^2} du, which is a simple power rule integration. This is a direct and effective application of the substitution.

Question 19

Let F(x)F(x) be defined by the integral F(x)=∫2exln⁡(t)tdtF(x) = \int_2^{e^x} \frac{\ln(t)}{t} dt What is the value of F′(ln⁡3)F'(\ln 3)?

  1. ln⁡3\ln 3 (correct answer)
  2. ln⁡33\frac{\ln 3}{3}
  3. 33
  4. 3ln⁡33 \ln 3
Explanation: This problem requires the Fundamental Theorem of Calculus, Part 2, combined with the Chain Rule. If F(x)=∫ah(x)f(t)dtF(x) = \int_a^{h(x)} f(t) dt, then F′(x)=f(h(x))⋅h′(x)F'(x) = f(h(x)) \cdot h'(x). In this case, f(t)=ln⁡(t)tf(t) = \frac{\ln(t)}{t} and the upper limit is h(x)=exh(x) = e^x. The derivative of the upper limit is h′(x)=exh'(x) = e^x. First, find F′(x)F'(x): F′(x)=f(ex)⋅ex=ln⁡(ex)ex⋅ex=xex⋅ex=xF'(x) = f(e^x) \cdot e^x = \frac{\ln(e^x)}{e^x} \cdot e^x = \frac{x}{e^x} \cdot e^x = x Now, evaluate F′(x)F'(x) at x=ln⁡3x = \ln 3: F′(ln⁡3)=ln⁡3F'(\ln 3) = \ln 3 The u-substitution skill is implicitly tested by recognizing the structure of the integrand f(t)=ln⁡ttf(t) = \frac{\ln t}{t} as a composite function suitable for integration, which informs the application of the FTC.

Question 20

A particle moves along a line with velocity v(t)=tt2+1v(t) = \frac{t}{t^2+1} meters per second at time t≥0t \ge 0. What is the total distance traveled by the particle from t=0t=0 to t=3t=3 seconds?

  1. ln⁡(10)\ln(10)
  2. 12ln⁡(10)\frac{1}{2} \ln(10) (correct answer)
  3. arctan⁡(3)\arctan(3)
  4. 910\frac{9}{10}
Explanation: When you see a velocity function and need to find total distance traveled, you're working with the fundamental relationship between velocity, displacement, and distance. The key insight is that distance is always positive, while displacement can be negative if the particle changes direction. To find total distance, you need to integrate the absolute value of velocity: ∫03∣v(t)∣ dt\int_0^3 |v(t)| \, dt. However, first check if the velocity changes sign in the interval. Since v(t)=tt2+1v(t) = \frac{t}{t^2+1} and both t≥0t \geq 0 and t2+1>0t^2+1 > 0 for t∈[0,3]t \in [0,3], the velocity is always non-negative. This means ∣v(t)∣=v(t)|v(t)| = v(t), so you can integrate directly. Calculate ∫03tt2+1 dt\int_0^3 \frac{t}{t^2+1} \, dt. Using substitution with u=t2+1u = t^2+1, so du=2t dtdu = 2t \, dt and t dt=12dut \, dt = \frac{1}{2} du. When t=0,u=1t=0,u=1; when t=3,u=10t=3,u=10. The integral becomes 12∫1101u du=12[ln⁡(u)]110=12(ln⁡(10)−ln⁡(1))=12ln⁡(10)\frac{1}{2} \int_1^{10} \frac{1}{u} \, du = \frac{1}{2}[\ln(u)]_1^{10} = \frac{1}{2}(\ln(10) - \ln(1)) = \frac{1}{2}\ln(10). Choice A) ln⁡(10)\ln(10) results from forgetting the 12\frac{1}{2} factor in the substitution. Choice C) arctan⁡(3)\arctan(3) comes from incorrectly integrating 1t2+1\frac{1}{t^2+1} instead of tt2+1\frac{t}{t^2+1}. Choice D) 910\frac{9}{10} suggests a computational error or wrong approach entirely. Remember: always check if velocity changes sign when finding distance traveled, and be careful with substitution factors during integration.