📝 Evaluating Definite Integrals by Substitution (25 MCQs)

📖 Explanation: Let $u = \sin x$, so $du = \cos x dx$. When $x=0, u=0$; when $x=\pi/2, u=1$. The integral becomes $\int_{0}^{1} u^3 du = [u^4/4]_{0}^{1} = 1/4$. This is a direct Easy of the substitution method for definite integrals, where changing the limits of integration is crucial to avoid back-substitution.

📖 Explanation: The substitution $u = e^x$ would make the integral more complex because $du = e^x dx$ and $x = \ln u$, leading to $\int \frac{\ln(\ln u)}{u} du$. The other substitutions simplify the integrand. $u = \ln x$ gives $du = dx/x$. $u = 1/x$ gives $du = -dx/x^2$, leading to $-\int u \ln(1/u) du$. $u = (\ln x)^2/2$ is an antiderivative form.

📖 Explanation: With $u=9-x^2$, $du=-2x dx$, so $x dx = -du/2$. Limits: $x=0 \to u=9$, $x=4 \to u=-7$. The integral becomes $-\frac{1}{2}\int_{9}^{-7} \sqrt{u} du$. The student might forget the negative sign or incorrectly transform the limits, leading to common errors like option C which has incorrect limits and sign.

📖 Explanation: Let $u = 2x$, then $du = 2dx$, so $dx = du/2$. Limits: $x=0 \to u=0$, $x=1 \to u=2$. The integral becomes $\frac{1}{2}\int_{0}^{2} f(u) du = \frac{1}{2}(5) = 2.5$. This tests the understanding that substitution changes the limits and scales the integral, which is a common Easy of the method.

📖 Explanation: The integral $\int_{0}^{1} x e^{x^2} dx$ requires a non-linear substitution $u=x^2$. The others only require linear substitutions like $u=3x+2$, $u=2x$, or $u=4x-1$. This question distinguishes between recognizing the need for a linear vs. a more complex substitution, testing Medium of the method.

📖 Explanation: Let $u = \sin x$, $du = \cos x dx$. Limits: $x=-\pi/2 \to u=-1$, $x=\pi/2 \to u=1$. The integral becomes $\int_{-1}^{1} u^2 du = [u^3/3]_{-1}^{1} = 1/3 - (-1/3) = 2/3$. Since the integrand is even, the result is twice the integral from 0 to 1, which is $2(1/3)=2/3$.

📖 Explanation: The most frequent error is forgetting to change the limits of integration from $x$-values to corresponding $u$-values. Students often compute the antiderivative in terms of $u$, evaluate at the original $x$-limits, or forget to back-substitute for the final evaluation. This error leads to incorrect numerical answers and is a common misconception.

📖 Explanation: Let $u = x^2+1$, $du = 2x dx$, so $x dx = du/2$. Limits: $x=0 \to u=1$, $x=1 \to u=2$. The integral becomes $\frac{1}{2}\int_{1}^{2} f(u) du$. We are given $\int_{1}^{4} f(u) du = 7$, but the limit is 2, not 4, so we cannot determine the exact value. Wait, the question says $\int_{1}^{4} f(u) du = 7$. The new limits are 1 to 2, not 1 to 4. So the value is $\frac{1}{2}\int_{1}^{2} f(u) du$, which is not determined by the given information. Correct answer: Cannot be determined.

📖 Explanation: Let $u=x^2$, then $du=2x dx$, so $x dx = du/2$. The integrand $x^3\sqrt{x^2+1} dx = x^2 \sqrt{x^2+1} (x dx) = u \sqrt{u+1} (du/2)$. Limits: $x=0 \to u=0$, $x=2 \to u=4$. So the integral is $\frac{1}{2}\int_{0}^{4} u \sqrt{u+1} du$. The challenge is correctly identifying $x^3 dx$ as $x^2 (x dx)$.

📖 Explanation: Let $u=1-x^2$, $du=-2x dx$, so $x dx = -du/2$. Limits: $x=0 \to u=1$, $x=1 \to u=0$. The integral becomes $-\frac{1}{2}\int_{1}^{0} u^{-1/2} du = \frac{1}{2}\int_{0}^{1} u^{-1/2} du = [\sqrt{u}]_{0}^{1} = 1$. Or directly, the antiderivative is $-\sqrt{1-x^2}$, evaluated from 0 to 1 gives $-0 - (-1) = 1$. Option B correctly states the antiderivative and the value.

📖 Explanation: Let $u=x^2$, $du=2x dx$, so $x dx = du/2$. Limits: $x=a \to u=a^2$, $x=b \to u=b^2$. The integral becomes $\frac{1}{2}\int_{a^2}^{b^2} f(u) du = \frac{1}{2}[F(b^2)-F(a^2)]$. This demonstrates the chain rule in reverse and the change of limits, requiring an understanding that the argument of $F$ must match the new limits of integration.

📖 Explanation: Let $u = \tan x$, then $du = \sec^2 x dx$. Limits: $x=0 \to u=0$, $x=\pi/4 \to u=1$. The integral becomes $\int_{0}^{1} u du = [u^2/2]_{0}^{1} = 1/2$. This is a straightforward Easy where the derivative of the chosen $u$ is present in the integrand, making the substitution obvious and the integration simple.

📖 Explanation: Let $u = \ln x$, $du = dx/x$. Limits: $x=1 \to u=0$, $x=2 \to u=\ln 2$. The integral becomes $\int_{0}^{\ln 2} \frac{du}{u} = [\ln |u|]_{0}^{\ln 2}$. However, $\ln(0)$ is undefined, so the integral is improper. The correct evaluation would be $\lim_{\epsilon \to 0^+} \int_{\epsilon}^{\ln 2} \frac{du}{u} = \infty$. The student likely ignored the singularity at $x=1$. This is a common error in analyzing the domain of the integrand.

📖 Explanation: All steps are correct. Let $u=1-x$, $du=-dx$, $x=1-u$. Limits: $x=0 \to u=1$, $x=1 \to u=0$. So $\int_{0}^{1} x(1-x)^3 dx = \int_{1}^{0} (1-u)u^3 (-du) = -\int_{1}^{0} (1-u)u^3 du = \int_{0}^{1} (1-u)u^3 du$. This is correct. The question asks which is NOT a valid step, but all are valid, implying the student must recognize the equivalence.

📖 Explanation: Let $u = 1-x$, $du = -dx$. Limits: $x=0 \to u=1$, $x=1 \to u=0$. The integral becomes $\int_{1}^{0} f(u) (-du) = \int_{0}^{1} f(u) du = 3$. This shows that the integral is invariant under the substitution $x \to 1-x$. It tests the understanding that the value of the integral does not depend on the variable name and how substitution affects the limits.

📖 Explanation: The function $f(x) = x^3 \sqrt{1-x^2}$ is odd because $f(-x) = -x^3 \sqrt{1-x^2} = -f(x)$. The interval $[-1,1]$ is symmetric about the origin. The definite integral of an odd function over a symmetric interval is 0. This can also be checked by substitution $u=1-x^2$ or $x = \sin \theta$, but the symmetry argument is most efficient. This is a classic Easy reasoning question.

📖 Explanation: Let $u = \cos x$, $du = -\sin x dx$. Limits: $x=0 \to u=1$, $x=\pi \to u=-1$. The integral becomes $-\int_{1}^{-1} \cos u du = \int_{-1}^{1} \cos u du = [\sin u]_{-1}^{1} = \sin(1) - \sin(-1) = 2\sin(1)$. Wait, the correct answer is $2\sin(1)$. But option B is $2\sin(1)$. Let's check: $-\int_{1}^{-1} \cos u du = \int_{-1}^{1} \cos u du = [\sin u]_{-1}^{1} = \sin 1 - (-\sin 1) = 2\sin 1$. So the correct answer is B.

📖 Explanation: With $u=x+1$, $du=dx$, and $x=u-1$. Limits: $x=0 \to u=1$, $x=2 \to u=3$. The integral becomes $\int_{1}^{3} (u-1)\sqrt{u} du$. This is a correct Easy where the student must express $x$ in terms of $u$ and change the limits. Option A is correct. Common errors include not changing limits (B) or incorrectly expressing $x$ (C).

📖 Explanation: Let $u = 3x$, $du = 3 dx$, so $dx = du/3$. Then $\int \frac{du/3}{\sqrt{1-u^2}} = \frac{1}{3}\int \frac{du}{\sqrt{1-u^2}} = \frac{1}{3}\sin^{-1}(u) + C = \frac{1}{3}\sin^{-1}(3x) + C$. This requires recognizing the standard integral $\int \frac{du}{\sqrt{1-u^2}} = \sin^{-1} u + C$ and correctly handling the coefficient of $x$.

📖 Explanation: Let $u = e^x+1$, $du = e^x dx$. Limits: $x=0 \to u=2$, $x=1 \to u=e+1$. The integral becomes $\int_{2}^{e+1} \frac{du}{u} = [\ln u]_{2}^{e+1} = \ln(e+1) - \ln 2 = \ln(\frac{e+1}{2})$. This tests the substitution $u$ and the properties of logarithms. Option B is correct. A common error is to incorrectly evaluate $\ln(e+1)$ as $1+\ln 1$, leading to option A or C.

📖 Explanation: The average value is $\frac{1}{1-0}\int_{0}^{1} x e^{x^2} dx$. Let $u=x^2$, $du=2x dx$, so $x dx = du/2$. Limits: 0 to 1 become 0 to 1. The integral becomes $\frac{1}{2}\int_{0}^{1} e^u du = \frac{1}{2}(e-1)$. The average value is $\frac{e-1}{2}$. This combines the concept of average value with the substitution method, making it a Easy problem.

📖 Explanation: Let $u = x^2$, $du = 2x dx$, so $x dx = du/2$. Limits: $x=0 \to u=0$, $x=1 \to u=1$. The integral becomes $\frac{1}{2}\int_{0}^{1} f(u) du = \frac{1}{2}(4) = 2$. This is a straightforward Easy of substitution, but the trick is recognizing that the limits remain 0 to 1 and that the integral of $f(u)$ from 0 to 1 is given. This is a classic problem that tests the understanding that the variable of integration is a dummy variable.

📖 Explanation: Both $u = \sin x$ and $u = \cos x$ would work, but $u = \sin x$ is more direct because $du = \cos x dx$, and the remaining $\cos^4 x$ can be expressed as $(1-\sin^2 x)^2 = (1-u^2)^2$. If $u = \cos x$, $du = -\sin x dx$, and the remaining $\sin^3 x$ is not a simple power of $u$ without introducing a square root. $u = \sin x$ leads to $\int u^4 (1-u^2)^2 du$, which is a polynomial.

📖 Explanation: Let $u = x^4+1$, $du = 4x^3 dx$, so $x^3 dx = du/4$. Limits: $x=0 \to u=1$, $x=1 \to u=2$. The integral becomes $\frac{1}{4}\int_{1}^{2} u^{-1/2} du = \frac{1}{4}[2u^{1/2}]_{1}^{2} = \frac{1}{2}(\sqrt{2}-1)$. This is a non-linear substitution where the derivative of $u$ is exactly present in the integrand. Option B is correct. Option A is missing the factor of 1/2.

📖 Explanation: Let $u=4-x^2$, $du=-2x dx$, so $x dx = -du/2$. Limits: $x=0 \to u=4$, $x=2 \to u=0$. The integral becomes $-\frac{1}{2}\int_{4}^{0} \sqrt{u} du = \frac{1}{2}\int_{0}^{4} u^{1/2} du = \frac{1}{2}[\frac{2}{3}u^{3/2}]_{0}^{4} = \frac{1}{3}(8) = \frac{8}{3}$. The student's answer is correct. This question tests the student's ability to verify an answer and identify that the negative sign and changing limits are handled correctly.