📝 Repeated integration by parts examples (34 MCQs)

📖 Explanation: Repeated integration by parts requires differentiating $x^3$ successively until zero. The first differentiation gives $3x^2$, second gives $6x$, third gives 6, and fourth gives 0. Integrating $e^{2x}$ repeatedly gives $\frac{e^{2x}}{2}, \frac{e^{2x}}{4}, \frac{e^{2x}}{8}, \frac{e^{2x}}{16}$. Alternating signs starting with positive, the result is $e^{2x}(\frac{x^3}{2} - \frac{3x^2}{4} + \frac{3x}{4} - \frac{3}{8}) + C$. Common errors include incorrect signs or coefficients in the pattern.

📖 Explanation: Tabular integration organizes repeated differentiations of the polynomial $x^2 - 3x + 5$ (giving $2x - 3$, then 2, then 0) and repeated integrations of $\sin x$ (giving $-\cos x$, $-\sin x$, $\cos x$). With alternating signs $+, -, +$, the product-sum yields $-(x^2 - 3x + 5)\cos x + (2x - 3)\sin x + 2\cos x + C$. A common mistake is forgetting the alternating sign pattern or the final integration of the constant.

📖 Explanation: The polynomial $x^4$ requires differentiation 5 times to reach zero: $x^4 \to 4x^3 \to 12x^2 \to 24x \to 24 \to 0$. The table includes each derivative row, starting with the original function, so 5 rows are needed. This is a critical concept because the number of rows determines the number of terms in the final antiderivative. A common error is counting only until the first zero derivative, which would be 4 differentiations but 5 rows including the original.

📖 Explanation: The correct antiderivative is $-x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + C$. The student's answer $-x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + C$ is actually correct. Wait, re-evaluation shows the student's answer is correct. The Medium should identify that the student got it right. A common misconception is that the signs should alternate starting with negative, but for $e^{-x}$, the integral is $-e^{-x}$, which already includes a negative sign, leading to the pattern $-, +, -$. The student's answer matches this pattern. This question tests whether students can verify correctness by differentiation.

📖 Explanation: Repeated integration by parts is most efficient for integrals of the form $\int p(x) f(x) \, dx$ where $p(x)$ is a polynomial and $f(x)$ is a function that can be repeatedly integrated (like exponentials, sines, cosines). $\int x^5 e^{3x} \, dx$ fits this perfectly. $\int \ln x \, dx$ is a single Easy; $\int e^x \sin x \, dx$ requires a cyclic pattern; and $\int \frac{1}{x^2+1} \, dx$ is a basic arctangent integral. Recognizing the structure is key to choosing the right technique.

📖 Explanation: Differentiate $x^3 - 2x$ repeatedly: $3x^2 - 2$, $6x$, 6, 0. Integrate $\cos 2x$ repeatedly: $\frac{1}{2}\sin 2x$, $-\frac{1}{4}\cos 2x$, $-\frac{1}{8}\sin 2x$, $\frac{1}{16}\cos 2x$. Alternating signs yield the result in option A. Common mistakes include incorrect coefficients from integration of $\cos 2x$ or misapplying the sign pattern. Option B has a sign error in the second term, option C has signs reversed, and option D has an incorrect sign pattern.

📖 Explanation: Applying integration by parts once reduces the power of $x$ from $n$ to $n-1$. This process continues until the power becomes 0. Therefore, exactly $n$ Easys are needed to eliminate the polynomial factor entirely. This is a fundamental understanding of repeated integration by parts. A common misconception is to think $n+1$ Easys are needed, confusing the number of terms in the final answer with the number of Easys. Each Easy reduces the exponent by exactly 1.

📖 Explanation: The correct result is $-x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6\sin x + C$. The student's answer has the first two terms correct, but the third and fourth terms show sign errors. In tabular integration, the signs alternate $+, -, +, -, \ldots$. For the third product, the sign should be $+$, giving $+6x \cos x$, and for the fourth product, the sign should be $-$, giving $-6\sin x$. The student likely alternated signs starting with $-$ for the first product, which is incorrect; the first product always has a positive sign. This is a common error in applying the tabular method.

📖 Explanation: The definite integral is found by evaluating the antiderivative at the limits 1 and 0. At $x=1$, the expression inside the parentheses is $-\frac{1}{2} - 1 - \frac{3}{2} - \frac{3}{2} - \frac{3}{4}$. At $x=0$, the polynomial part $-x^4/2 - x^3 - 3x^2/2 - 3x/2 - 3/4$ evaluates to $-3/4$. So the definite integral is $e^{-2}(-\frac{1}{2} - 1 - \frac{3}{2} - \frac{3}{2} - \frac{3}{4}) - (-3/4) = e^{-2}(-\frac{1}{2} - 1 - \frac{3}{2} - \frac{3}{2} - \frac{3}{4}) + \frac{3}{4}$. Common mistakes include forgetting to evaluate at the lower limit or sign errors.

📖 Explanation: In tabular integration by parts, the alternating signs are applied to each product formed from a row of the polynomial's derivatives and the corresponding row of the second function's integrals. Starting with a positive sign for the first product, signs alternate for each subsequent product. This is a crucial part of the method. A common misconception is that the signs are related to the derivatives of the polynomial or the integrals of the second function, but they are independent and simply part of the integration by parts formula. The pattern comes directly from the $(-1)^n$ factor in the repeated Easy of the integration by parts formula.

📖 Explanation: Differentiate $3x^2 + 2x - 1$ repeatedly: $6x + 2$, 6, 0. Integrate $e^x$ repeatedly: $e^x, e^x, e^x$. Products with alternating signs: $+(3x^2+2x-1)e^x - (6x+2)e^x + 6e^x = e^x(3x^2+2x-1 -6x-2+6) = e^x(3x^2 -4x+3)$. Wait, the last term should be $+6e^x$, so $-1 -2 +6 = 3$. The result is $e^x(3x^2 -4x+3)$. Option B is $e^x(3x^2 -4x+5)$, option C is $e^x(3x^2+8x+7)$, option D is $e^x(3x^2 -4x+7)$. The correct answer is $e^x(3x^2 -4x+3)$. I'll adjust options. Let's recompute: $(3x^2+2x-1) - (6x+2) + 6 = 3x^2 +2x -1 -6x -2 +6 = 3x^2 -4x +3$. So correct answer is $e^x(3x^2 -4x+3)$.

📖 Explanation: Tabular integration by parts is specifically designed for integrals of the form $\int p(x)f(x) \, dx$ where $p(x)$ is a polynomial and eventually differentiates to zero. For $\int e^x \cos x \, dx$, neither $e^x$ nor $\cos x$ differentiates to zero; the derivatives cycle. Therefore, the tabular method as described (differentiating until zero) would never terminate. A cyclic method must be used instead. This highlights the limitation of tabular integration. A common error is to apply tabular integration to any product, but it's only efficient when one factor is a polynomial.

📖 Explanation: This is a classic example where tabular integration is effective. Differentiate $x^2$ repeatedly: $2x$, 2, 0. Integrate $\sqrt{x-1} = (x-1)^{1/2}$ repeatedly: $\frac{2}{3}(x-1)^{3/2}$, $\frac{4}{15}(x-1)^{5/2}$, $\frac{8}{105}(x-1)^{7/2}$. Products with alternating signs: $+\frac{2}{3}x^2(x-1)^{3/2} - \frac{8}{15}x(x-1)^{5/2} + \frac{16}{105}(x-1)^{7/2} + C$. Option B has signs, option C has a plus in the second term, and option D has a minus in the last term. The correct result is in option A. This integral also illustrates that tabular integration can be used even when the second function involves a fractional power.

📖 Explanation: Tabular integration by parts requires that the second function can be repeatedly integrated in a closed form. While $x^3$ differentiates to zero, $e^{x^2}$ does not have an elementary antiderivative, so its repeated integrals are not expressible in terms of elementary functions. Thus, tabular integration by parts is not applicable. A common misconception is that tabular integration by parts works whenever the polynomial differentiates to zero, but the other function must be repeatedly integrable. The correct approach for this integral is a substitution $u = x^2$, which simplifies it to $\frac{1}{2} \int u e^u \, du$, which then requires a single integration by parts.

📖 Explanation: While tabular integration by parts is efficient for polynomial times exponentials or trigonometric functions, $\ln x$ is a logarithmic function. The LIATE rule suggests choosing $u = \ln x$ (logarithmic) and $dv = x^5 dx$ (algebraic). This requires only one integration by parts: $\int x^5 \ln x \, dx = \frac{x^6}{6}\ln x - \int \frac{x^6}{6} \cdot \frac{1}{x} dx = \frac{x^6}{6}\ln x - \frac{x^6}{36} + C$. Tabular integration would not be efficient here because differentiating $\ln x$ gives $1/x$, which doesn't reduce to a polynomial, and integrating $x^5$ repeatedly would not terminate nicely. This question tests the ability to choose the most appropriate technique based on the integrand's structure.

📖 Explanation: The polynomial $x^3$ requires 4 differentiations to reach zero (including the original function): $x^3, 3x^2, 6x, 6, 0$. Each non-zero row contributes a term to the final antiderivative. Therefore, there will be 4 terms in the result. The first term involves $x^3$, the second $x^2$, the third $x$, and the fourth is a constant multiplied by a sine or cosine. A common mistake is to forget the constant term that results from integrating the last non-zero derivative, leading to only 3 terms. The number of terms equals the degree of the polynomial plus 1.

📖 Explanation: Differentiate the polynomial: $3x^2+4x-1$, $6x+4$, 6, 0. Integrate $e^{-x}$: $-e^{-x}, e^{-x}, -e^{-x}, e^{-x}$. Products with alternating signs: $+(x^3+2x^2-x+1)(-e^{-x}) - (3x^2+4x-1)(e^{-x}) + (6x+4)(-e^{-x}) - 6(e^{-x})$. Simplify: $e^{-x}[-(x^3+2x^2-x+1) - (3x^2+4x-1) - (6x+4) - 6] = e^{-x}[-x^3 -2x^2 + x -1 -3x^2 -4x +1 -6x -4 -6] = e^{-x}[-x^3 -5x^2 -9x -10]$. The correct answer is $e^{-x}(-x^3 - 5x^2 - 9x - 10) + C$. Common errors include sign mistakes from integrating $e^{-x}$ or in the alternating pattern.

📖 Explanation: Differentiating $x^2\sin x + 2x\cos x - 2\sin x + C$: $d/dx(x^2\sin x) = 2x\sin x + x^2\cos x$. $d/dx(2x\cos x) = 2\cos x - 2x\sin x$. $d/dx(-2\sin x) = -2\cos x$. Sum: $x^2\cos x + 2x\sin x - 2x\sin x + 2\cos x - 2\cos x = x^2\cos x$. The derivative is exactly the integrand, so the student's answer is correct. This question teaches students to verify their antiderivatives by differentiation, a crucial skill. It also shows that the student's answer, which might look suspicious, is actually correct. A common error is to not verify or to make arithmetic mistakes during differentiation.

📖 Explanation: The tabular method organizes the repeated differentiations and integrations in a clear, visual format. This significantly reduces the chance of sign errors and makes the process less tedious, especially for high-degree polynomials. While it does reduce the number of steps in a sense, the primary advantage is the organized structure that minimizes mistakes. A common misconception is that the tabular method is a fundamentally different technique; it's actually just a streamlined way to apply integration by parts repeatedly. It also doesn't handle any integrand, only those where one factor is a polynomial that eventually differentiates to zero.

📖 Explanation: When integrating a polynomial times $\cos 3x$, the repeated integrations of $\cos 3x$ alternate between sine and cosine: $\frac{1}{3}\sin 3x$, $-\frac{1}{9}\cos 3x$, $-\frac{1}{27}\sin 3x$, $\frac{1}{81}\cos 3x$, etc. Since the polynomial has four differentiations (including zero), the final product will include both sine and cosine terms depending on the parity of the differentiation. For even differentiations, the integral is $\cos 3x$ or $-\cos 3x$; for odd differentiations, it's $\sin 3x$ or $-\sin 3x$. Thus, both functions appear. A common error is to assume only one trigonometric function appears because the integrand only has cosine.

📖 Explanation: When the polynomial differentiates to a constant, the tabular method uses that constant in the product for the last row, and then the integration of that constant is what gives the final term. If a student forgets this final term, their antiderivative will be missing a term that is a constant times the integral of the second function. For example, for $\int x^2 e^x dx$, forgetting the final term would give $x^2 e^x - 2x e^x$ instead of $x^2 e^x - 2x e^x + 2e^x$. This is a significant error because the derivative of the incorrect answer would be $x^2 e^x$, which is not the original integrand. The missing term is not the constant of integration; it's a functional term involving the integral of the second function.

📖 Explanation: Differentiate polynomial: $3x^2-4x$, $6x-4$, 6, 0. Integrate $\sin 2x$: $-\frac{1}{2}\cos 2x$, $-\frac{1}{4}\sin 2x$, $\frac{1}{8}\cos 2x$, $\frac{1}{16}\sin 2x$. Products with alternating signs: $+(x^3-2x^2+3)(-\frac{1}{2}\cos 2x) - (3x^2-4x)(-\frac{1}{4}\sin 2x) + (6x-4)(\frac{1}{8}\cos 2x) - 6(\frac{1}{16}\sin 2x)$. Simplify: $-\frac{1}{2}(x^3-2x^2+3)\cos 2x + \frac{1}{4}(3x^2-4x)\sin 2x + \frac{3}{4}(x - \frac{2}{3})\cos 2x - \frac{3}{8}\sin 2x$. Wait, the third term is $(6x-4)/8 = (3x-2)/4 = \frac{3}{4}(x - \frac{2}{3})$. So the result is $-\frac{1}{2}(x^3-2x^2+3)\cos 2x + \frac{1}{4}(3x^2-4x)\sin 2x + \frac{3}{4}(x - \frac{2}{3})\cos 2x - \frac{3}{8}\sin 2x + C$. Option B is correct. I'll adjust options. The correct answer is $-\frac{1}{2}(x^3-2x^2+3)\cos 2x + \frac{1}{4}(3x^2-4x)\sin 2x + \frac{3}{4}(x - \frac{2}{3})\cos 2x - \frac{3}{8}\sin 2x + C$.

📖 Explanation: The tabular method pairs the first derivative (which is the original function) with the first integral of $f(x)$, the second derivative with the second integral, and so on, moving down both columns simultaneously. This is the fundamental construction of the table. Pairing them in reverse order or arbitrarily would change the products and the result, and is not the correct procedure. A common misconception is that the columns can be paired in any order, but the method relies on a specific pairing that corresponds to the repeated Easy of the integration by parts formula. This question tests the procedural knowledge of the tabular method.

📖 Explanation: The antiderivative is $e^{-x}(-x^3 - 3x^2 - 6x - 6) + C$. Evaluate at 1: $e^{-1}(-1 - 3 - 6 - 6) = -16/e$. At 0: $e^0(0 - 0 - 0 - 6) = -6$. Definite integral = $(-16/e) - (-6) = 6 - 16/e$. This is the correct answer. Common mistakes include errors in the antiderivative, particularly sign errors, or mistakes in evaluating at the limits. Option C and D have different constants or signs. This problem combines tabular integration with definite integration, testing both skills.

📖 Explanation: The number of rows in the tabular table is determined by how many times $p(x)$ must be differentiated to reach zero, plus the original row. For $p(x) = x^3$, we have rows: $x^3$, $3x^2$, $6x$, 6, 0. That's 5 rows including the zero row, but typically the zero row is not listed as a separate row in the product formation, so we have 4 non-zero rows. For $p(x) = x^2$, rows: $x^2$, $2x$, 2, 0. That's 3 non-zero rows. For $p(x) = x^4$, rows: $x^4$, $4x^3$, $12x^2$, $24x$, 24, 0. That's 5 non-zero rows. So $p(x) = x^3$ has 4 non-zero rows. A common mistake is to count the zero row or miscount the differentiations.

📖 Explanation: The antiderivative of $t^2 e^{-kt}$ is $e^{-kt}(-\frac{t^2}{k} - \frac{2t}{k^2} - \frac{2}{k^3}) + C$. Evaluating from 0 to T: at T, $e^{-kT}(-\frac{T^2}{k} - \frac{2T}{k^2} - \frac{2}{k^3})$. At 0, the expression is $-\frac{2}{k^3}$. The definite integral is $e^{-kT}(-\frac{T^2}{k} - \frac{2T}{k^2} - \frac{2}{k^3}) - (-\frac{2}{k^3}) = \frac{2}{k^3} - e^{-kT}(\frac{T^2}{k} + \frac{2T}{k^2} + \frac{2}{k^3})$. This is a common result in probability and physics. A common error is forgetting the lower limit or sign errors. This question connects calculus to a real-world Easy, making it more relevant and Hard.

📖 Explanation: The antiderivative will be of the form $e^{ax} \cdot P(x)$, where $P(x)$ is a polynomial. Since $a > 0$, as $x \to \infty$, the exponential $e^{ax}$ dominates any polynomial, so the antiderivative grows without bound, i.e., it grows exponentially. This is a key concept in understanding the behavior of integrals involving exponentials. A common misconception is that the polynomial part dominates because it's of high degree, but exponential growth always outpaces polynomial growth. This question requires understanding the asymptotic behavior of the antiderivative based on its form.

📖 Explanation: Tabular integration by parts requires that the non-polynomial function can be repeatedly integrated in a closed form (i.e., an elementary antiderivative exists for each integration). While $\sec x$ has an elementary antiderivative ($\ln|\sec x + \tan x|$), its repeated integrals do not generally yield elementary functions. Thus, the table would not terminate in a usable form. This is a common pitfall. Students often think any function can be used, but $\sec x$, $\tan x$, and many others do not have simple repeated integrals. The correct approach for such integrals often involves different techniques or recognition that they might not have elementary antiderivatives.

📖 Explanation: Differentiate $x^3 - 2x$: $3x^2 - 2$, $6x$, 6, 0. Integrate $e^{-x}$: $-e^{-x}, e^{-x}, -e^{-x}, e^{-x}$. Products: $+(x^3-2x)(-e^{-x}) - (3x^2-2)(e^{-x}) + (6x)(-e^{-x}) - 6(e^{-x})$. Simplify: $e^{-x}[-x^3+2x - 3x^2 +2 -6x -6] = e^{-x}[-x^3 -3x^2 -4x -4]$. The coefficient of $x^2$ is -3. I need to adjust options. Let's recompute the $x^2$ coefficient: from the second product $-(3x^2-2)$, we get $-3x^2$. So the coefficient is -3. The correct answer is -3.

📖 Explanation: Products with alternating signs: $+x^4(\sin x) - 4x^3(-\cos x) + 12x^2(-\sin x) - 24x(\cos x) + 24(\sin x) = x^4\sin x + 4x^3\cos x - 12x^2\sin x - 24x\cos x + 24\sin x + C$. This is option A. Option B has the same expression. I'll make the options distinct. This question tests the ability to correctly apply the tabular method given a partially completed table. It requires understanding of alternating signs and the pairing of derivatives and integrals. Common errors include sign mistakes or incorrect pairing.

📖 Explanation: This is an improper integral. First find the antiderivative: $e^{-2x}(-\frac{x^3}{2} - \frac{3x^2}{4} - \frac{3x}{4} - \frac{3}{8}) + C$. Evaluate from 0 to b: $e^{-2b}(-\frac{b^3}{2} - \frac{3b^2}{4} - \frac{3b}{4} - \frac{3}{8}) - (-\frac{3}{8})$. As $b \to \infty$, the exponential term goes to 0, so the limit is $\frac{3}{8}$. This is a known result: $\int_0^\infty x^n e^{-ax} dx = \frac{n!}{a^{n+1}}$. For n=3, a=2, it's $\frac{6}{16} = \frac{3}{8}$. This combines repeated integration by parts with improper integrals and limits, making it a Hard problem. A common error is forgetting the lower limit or making mistakes in the improper integral evaluation.

📖 Explanation: Repeatedly integrating $\cos(ax)$ alternates between $\cos$ and $\sin$ with powers of $1/a$. The polynomial differentiations produce polynomials of decreasing degree. The final antiderivative will be a linear combination of $\cos(ax)$ and $\sin(ax)$, each multiplied by a polynomial. The degree of these polynomials is at most $n$. This is a general result for such integrals. This question requires understanding the pattern that emerges from repeated integration by parts. A common misconception is that the result involves only one trigonometric function or that the polynomials have degree $n+1$. The correct answer is the one that captures both the polynomial and trigonometric nature of the result.

📖 Explanation: Let's compute the correct integral: $\int x^2 e^{3x} dx = e^{3x}(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}) + C = \frac{e^{3x}}{27}(9x^2 - 6x + 2) + C$. The student's answer $\frac{e^{3x}}{27}(9x^2 - 6x + 2)$ has the polynomial $9x^2 - 6x + 2$. This is actually correct! Wait, the student's answer is the same. Let's re-evaluate the student's answer: $\frac{e^{3x}}{27}(9x^2 - 6x + 2) = e^{3x}(\frac{9x^2}{27} - \frac{6x}{27} + \frac{2}{27}) = e^{3x}(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27})$. This is correct. The student made no mistake. The options need to be adjusted. Let's make the student's answer $\frac{e^{3x}}{27}(9x^2 + 6x + 2)$. Then the mistake is the sign of the $x$ term. The correct polynomial should have $-6x$. So the mistake is a sign error in the second term. This question highlights that students should verify their antiderivatives by differentiation.

📖 Explanation: The tabular method is most directly applicable when one of the functions is a polynomial (or can be differentiated to zero in a finite number of steps). While it can be adapted for other cases, the standard form relies on this condition. The integral can be definite or indefinite, the interval can be finite or infinite, and the integrand can be a product of more than two functions if grouped appropriately. This question tests the core prerequisite for using the tabular method effectively. A common misconception is that the method works for any product, but it's only efficient and straightforward when one factor is a polynomial.