Integrating Using Integration By Parts - AP Calculus BC

A signal uses $\int x\sin x\,dx$ . Which expression is a correct antiderivative using integration by parts?
Answer: $-x\cos x+\sin x+C$ This problem requires integration by parts to find the antiderivative of x sin x. To apply integration by parts, choose u = x because its derivative is 1, and dv = sin x dx because sine integrates to -cos x. Then du = dx and v = -cos x. Substituting into the formula gives -x cos x - ∫ -cos x dx, which simplifies to -x cos x + sin x + C. A tempting distractor might be x sin x + cos x + C, which could result from incorrect choice of v or sign errors. A transferable strategy for integration by parts is to assign u to the polynomial to reduce its power, and dv to the trigonometric function for straightforward integration.
In a probability derivation, compute $\int \ln(2x)\,dx$ for $x>0$ using integration by parts.
Answer: $x\ln(2x)-x+C$ This problem requires integration by parts to find the antiderivative of ln(2x) for x > 0. To apply integration by parts, choose u = ln(2x) because its derivative is 1/x, and dv = dx because it integrates to x. Then du = (1/x) dx and v = x. Substituting into the formula gives x ln(2x) - ∫ x (1/x) dx, which simplifies to x ln(2x) - x + C. A tempting distractor might be x ln(2x) + x + C, which could arise from mishandling the sign in the subtraction step. A transferable strategy for integration by parts is to let u be the logarithmic term for differentiation to a rational function, and dv the constant or simple term for easy integration.
A model uses $\int x^{2}e^{x}\,dx$ ; which expression is a correct antiderivative?
Answer: $e^{x}(x^{2}-2x+2)+C$ This integral requires integration by parts twice for $x^2 e^x$ . First, let $u = x^2$ (so $du = 2x \, dx$ ) and $dv = e^x \, dx$ (so $v = e^x$ ), giving $∫ x^2 e^x \, dx = x^2 e^x - ∫ 2x e^x \, dx$ . For the remaining integral $∫ 2x e^x \, dx$ , apply parts again with $u = 2x$ and $dv = e^x \, dx$ , yielding $2x e^x - 2 e^x$ . Combining: $∫ x^2 e^x \, dx = x^2 e^x - 2x e^x + 2 e^x + C = e^x (x^2 - 2x + 2) + C$ . Choice D shows only the first application of parts without completing the process. For polynomial times exponential, repeated integration by parts reduces the polynomial degree to zero.
A signal's accumulated energy is $\int e^{x}\cos x\,dx$ ; which is an antiderivative?
Answer: $\frac{e^{x}}{2}(\sin x+\cos x)+C$ This integral ∫ $e^x$ cos x dx requires integration by parts twice, creating a cyclic pattern. First, let u = cos x and dv = $e^x$ dx, giving ∫ $e^x$ cos x dx = $e^x$ cos x - ∫ $e^x$ (-sin x) dx = $e^x$ cos x + ∫ $e^x$ sin x dx. For the new integral, use u = sin x and dv = $e^x$ dx: ∫ $e^x$ sin x dx = $e^x$ sin x - ∫ $e^x$ cos x dx. This gives us back our original integral, so combining both equations and solving: 2∫ $e^x$ cos x dx = $e^x$ cos x + $e^x$ sin x, thus ∫ $e^x$ cos x dx = ( $e^x$ /2)(sin x + cos x) + C. Choice D shows the incomplete first step without recognizing the cyclic pattern. For exponential times trig functions, expect to use parts twice and solve algebraically.
A particle's momentum involves $\int x e^{2x}\,dx$ . Which expression is an antiderivative of $x e^{2x}$ ?
Answer: $\frac{x e^{2x}}{2}-\frac{e^{2x}}{4}+C$ This integral requires integration by parts with the exponential function $e^{2x}$ . Choose u = x (so du = dx) and dv = $e^{2x}$ dx, which gives v = $e^{2x}$ /2 after integrating. Applying the parts formula: ∫x $e^{2x}$ dx = x( $e^{2x}$ /2) - ∫( $e^{2x}$ /2) dx = (x $e^{2x}$ )/2 - ( $e^{2x}$ /4) + C. Option A incorrectly adds $e^{2x}$ /4 instead of subtracting it, a common sign error in the second term. When integrating products with exponentials $e^{ax}$ , remember that integrating $e^{ax}$ gives $e^{ax}$ /a, and always check your signs carefully in the final expression.
In a heat-transfer derivation, compute $\int x\ln x\,dx$ for $x>0$ .
Answer: $\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}+C$ Integration by parts is the key skill for evaluating ∫ x ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = x dx so v = (1/2) $x^2$ . This choice works well as ln x simplifies upon differentiation, pairing with the polynomial growth in dv. The result is (1/2) $x^2$ ln x - ∫ (1/2) $x^2$ (1/x) dx, simplifying to (1/2) $x^2$ ln x - (1/4) $x^2$ + C. A tempting distractor like choice B fails with a positive (1/4) $x^2$ term, likely from neglecting the subtraction in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
To compute average power, find $\int x\sec^2 x\,dx$ using integration by parts.
Answer: $x\tan x-\ln|\sec x|+C$ Integration by parts is the key skill for evaluating ∫ x $\sec^2$ x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = x so du = dx, and dv = $\sec^2$ x dx so v = tan x. This selection is effective because the polynomial x simplifies, leaving a trigonometric integral. The computation gives x tan x - ∫ tan x dx, which integrates to x tan x - ln|sec x| + C. A tempting distractor like choice B fails with a positive ln|sec x| term, possibly from inverting the sign when integrating tan x. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
For a cost model, evaluate $\int \ln x\,dx$ (for $x>0$ ) using integration by parts.
Answer: $x\ln x - x + C$ Integration by parts is the key skill for evaluating ∫ ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = dx so v = x. This selection is ideal because ln x differentiates to a simple rational function, while dv is basic. The result is x ln x - ∫ x (1/x) dx, simplifying to x ln x - x + C. A tempting distractor like choice B fails with a positive x term, which would result from forgetting the negative sign in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
While finding area, you need $\int x^2\ln x\,dx$ for $x>0$ . Which is an antiderivative?
Answer: $\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$ This problem requires integration by parts to find the antiderivative of $x^2$ ln x for x > 0. To apply integration by parts, choose u = ln x because it differentiates to 1/x, and dv = $x^2$ dx because it integrates to (1/3) $x^3$ . Then du = (1/x) dx and v = (1/3) $x^3$ . Substituting into the formula gives (1/3) $x^3$ ln x - ∫ (1/3) $x^3$ (1/x) dx, which simplifies to (1/3) $x^3$ ln x - (1/9) $x^3$ + C. A tempting distractor might be (1/3) $x^3$ ln x + (1/9) $x^3$ + C, stemming from a sign error in the parts formula. A transferable strategy for integration by parts is to select u as the logarithm to simplify, and dv as the higher-power polynomial to integrate.
The accumulated charge is $\int x\ln x\,dx$ for $x>0$ ; which is an antiderivative?
Answer: $\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C$ This integral requires integration by parts for the product x ln x. We must choose u = ln x (giving du = 1/x dx) and dv = x dx (giving v = x²/2), because the derivative of ln x is algebraic while its integral is more complex. Applying the formula: ∫x ln x dx = (ln x)(x²/2) - ∫(x²/2)(1/x) dx = (x²/2)ln x - ∫x/2 dx = (x²/2)ln x - x²/4 + C. Choice D shows the incomplete integration by parts setup, failing to evaluate the remaining integral. When integrating products involving ln x, always let u = ln x since its derivative simplifies to 1/x.
A signal's accumulated effect is $\int x^2\ln(x)\,dx$ . Which option is a correct antiderivative?
Answer: $\frac{x^3}{3}\ln(x)-\frac{x^3}{9}+C$ This integral combines a power function x² with ln(x), requiring integration by parts. Choose u = ln(x) (so du = (1/x) dx) and dv = x² dx (giving v = x³/3). Applying the parts formula: ∫x² ln(x) dx = (x³/3)ln(x) - ∫(x³/3)(1/x) dx = (x³/3)ln(x) - ∫(x²/3) dx = (x³/3)ln(x) - x³/9 + C. Option B incorrectly adds x³/9 instead of subtracting it, a sign error in applying the integration by parts formula. When integrating products of powers of x with ln(x), always let u = ln(x) to simplify the derivative, and carefully track the arithmetic when integrating the remaining power function.
A growth model uses $\int e^x\sin x\,dx$ ; choose the correct antiderivative.
Answer: $\tfrac{1}{2}e^x(\sin x-\cos x)+C$ Integration by parts is the key skill for evaluating ∫ $e^x$ sin x dx, based on the formula ∫ u dv = u v - ∫ v du, applied twice to cycle back. Choose u = sin x so du = cos x dx, and dv = $e^x$ dx so v = $e^x$ for the first step. This produces $e^x$ sin x - ∫ $e^x$ cos x dx, then repeat with u = cos x and dv = $e^x$ dx. Solving the system yields (1/2) $e^x$ (sin x - cos x) + C. A tempting distractor like choice B fails by having a positive cos x term, which might come from a sign error in resolving the cyclic integrals. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
A sensor's cumulative output is modeled by $\int x e^{x}\,dx$ ; which expression is an antiderivative?
Answer: $e^{x}(x-1)+C$ This integral requires integration by parts to handle the product of x and $e^x$ . For integration by parts, we use ∫u dv = uv - ∫v du, choosing u = x (so du = dx) and dv = $e^x$ dx (so v = $e^x$ ). Applying the formula: ∫ $xe^x$ dx = $xe^x$ - ∫ $e^x$ dx = $xe^x$ - $e^x$ + C = $e^x$ (x-1) + C. Choice C shows the incomplete form before finishing the integration, which is a common error when students forget to complete the final integral. When using integration by parts with polynomial times exponential, always let u be the polynomial part to reduce its degree.
In modeling work done, find an antiderivative of $\int x e^{2x}\,dx$ using integration by parts.
Answer: $\tfrac{1}{2}xe^{2x}-\tfrac{1}{4}e^{2x}+C$ Integration by parts is the key skill for evaluating ∫ x $e^{2x}$ dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = x so du = dx, and dv = $e^{2x}$ dx so v = (1/2) $e^{2x}$ . This selection works because differentiating x simplifies the integrand, while integrating the exponential is straightforward. Applying the formula yields (1/2) x $e^{2x}$ - ∫ (1/2) $e^{2x}$ dx, which simplifies to (1/2) x $e^{2x}$ - (1/4) $e^{2x}$ + C. A tempting distractor like choice B fails because it has a positive (1/4) $e^{2x}$ term, which would arise from an incorrect sign in the subtraction step. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
A particle's momentum term involves $x^2\sin x$ ; find $\int x^2\sin x\,dx$ .
Answer: $-x^2\cos x+2x\sin x+2\cos x+C$ Integration by parts is the key skill for evaluating ∫ $x^2$ sin x dx, based on the formula ∫ u dv = u v - ∫ v du, often applied repeatedly. Choose u = $x^2$ so du = 2x dx, and dv = sin x dx so v = -cos x for the first step. This leads to - $x^2$ cos x + 2 ∫ x cos x dx, requiring another parts application where u = x and dv = cos x dx. The full computation yields - $x^2$ cos x + 2x sin x + 2 cos x + C after integrating twice. A tempting distractor like choice C fails by having a negative 2 cos x term, possibly from a sign error in the second integration step. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
In a heat-transfer model, compute $\int \ln(x)\,dx$ . Which expression is an antiderivative of $\ln(x)$ ?
Answer: $x\ln(x)-x+C$ To integrate ln(x), we use integration by parts with a clever choice: u = ln(x) and dv = dx. This gives du = (1/x) dx and v = x. Applying the formula: ∫ln(x) dx = x ln(x) - ∫x(1/x) dx = x ln(x) - ∫1 dx = x ln(x) - x + C. Option A incorrectly adds x instead of subtracting it, missing the negative sign that comes from the integration by parts formula. For integrals of logarithms alone, always use dv = dx (essentially treating it as ln(x)·1) to convert the problem into a simpler form.
In a probability derivation, evaluate $\int x^3 e^x \, dx$ .
Answer: $e^x(x^3-3x^2+6x-6)+C$ Integration by parts is the key skill for evaluating $\int x^3 e^x \, dx$ , based on the formula $\int u \, dv = u v - \int v \, du$ , repeated for the polynomial. Choose $u = x^3$ so $du = 3x^2 \, dx$ , and $dv = e^x \, dx$ so $v = e^x$ , then repeat for lower powers. This tabular approach reduces the degree each time, leading to subsequent integrals of $x^2 e^x$ , $x e^x$ , and $e^x$ . The full expansion yields $e^x (x^3 - 3x^2 + 6x - 6) + C$ . A tempting distractor like choice B fails with all positive coefficients, likely from consistent sign errors in the repeated subtractions. A transferable strategy for integration by parts is to prioritize $u$ as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.
A temperature function includes $\int e^x\cos x\,dx$ . Which is an antiderivative obtained by integration by parts twice?
Answer: $\frac{1}{2}e^x(\cos x+\sin x)+C$ This problem requires integration by parts to find the antiderivative of $e^x$ cos x, applied twice. To apply integration by parts, choose u = cos x and dv = $e^x$ dx for the first step, as the exponential integrates to itself; then du = -sin x dx and v = $e^x$ . This leads to $e^x$ cos x + ∫ $e^x$ sin x dx; repeating with u = sin x and dv = $e^x$ dx gives $e^x$ sin x - ∫ $e^x$ cos x dx, resulting in the original integral on both sides and solving to (1/2) $e^x$ (cos x + sin x) + C. A tempting distractor might be without the 1/2 factor, from forgetting to divide after solving the equation. A transferable strategy for integration by parts is to alternate choices of u and dv for products of exponentials and trig functions, leading to a solvable equation when the integral reappears.
A damping model requires $\int x e^{-x}\,dx$ ; which expression is an antiderivative?
Answer: $-e^{-x}(x+1)+C$ This integral requires integration by parts for $xe^{-x}$ . Choose u = x (so du = dx) and dv = $e^{-x}$ dx (so v = - $e^{-x}$ ), noting the negative from integrating $e^{-x}$ . Applying the formula: ∫ $xe^{-x}$ dx = x(- $e^{-x}$ ) - ∫(- $e^{-x}$ ) dx = - $xe^{-x}$ + ∫ $e^{-x}$ dx = - $xe^{-x}$ - $e^{-x}$ + C = - $e^{-x}$ (x + 1) + C. Choice C has the wrong overall sign, missing the negative that comes from integrating $e^{-x}$ . When integrating products with $e^{-x}$ , remember that ∫ $e^{-x}$ dx = - $e^{-x}$ , introducing a crucial negative sign.
An object's work is modeled by $W=\int x\ln(x)\,dx$ . Which expression is an antiderivative of $x\ln(x)$ ?
Answer: $\frac{x^2}{2}\ln(x)-\frac{x^2}{4}+C$ This integral requires integration by parts to handle the product of x and ln(x). For integration by parts, we use the formula ∫u dv = uv - ∫v du, where we choose u = ln(x) (since its derivative 1/x is simpler) and dv = x dx (making v = x²/2). Applying the formula: ∫x ln(x) dx = (x²/2)ln(x) - ∫(x²/2)(1/x) dx = (x²/2)ln(x) - ∫(x/2) dx = (x²/2)ln(x) - x²/4 + C. Option B incorrectly adds x²/4 instead of subtracting it, which is a common sign error when applying integration by parts. When integrating products involving logarithms, always choose u = ln(x) since differentiating logarithms simplifies the expression while integrating them makes things more complex.

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