📝 Tabular integration by parts method (35 MCQs)

📖 Explanation: Tabular integration by parts is a systematic method for integrating products where one function (typically a polynomial) eventually differentiates to zero. In $\int (x^3 - 2x) \sin x \, dx$, the polynomial $x^3 - 2x$ is repeatedly differentiated until its derivative becomes 0. The trigonometric function $\sin x$ is integrated repeatedly in the adjacent column, creating a pattern that does not terminate on its own. The process stops because the polynomial column reaches zero, not because the trigonometric column terminates.

📖 Explanation: The LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) suggests that the Algebraic function $x^2$ should be chosen as $u$ (the function to differentiate) because it simplifies upon differentiation. The Exponential function $e^{-2x}$ should be chosen as $dv$ (the function to integrate) because it is easily integrated and does not become more complicated. This arrangement ensures that the derivatives of $x^2$ eventually reach zero, making tabular integration efficient. The other options violate LIATE principles and would lead to a more complicated or non-terminating process.

📖 Explanation: In tabular integration by parts, the process must continue until the polynomial column differentiates to zero. For $x^3$, the derivatives are $3x^2, 6x, 6, 0$. The student stopped at the third derivative (6) without completing the final row where the derivative becomes 0. The correct result should include the last term from the integration of the zero row, and the signs must follow the alternating pattern. Stopping prematurely leads to missing terms and an incorrect constant of integration. The student also likely misapplied the sign pattern, as the correct answer requires $+6\sin x$ from the final integration of the zero row.

📖 Explanation: Tabular integration by parts is most efficient when one function in the product is a polynomial that can be differentiated repeatedly to zero. For $\int x^4 e^x \, dx$, repeated differentiation of $x^4$ yields a finite number of terms (derivatives $4x^3, 12x^2, 24x, 24, 0$), making tabular integration a structured and efficient method. Standard integration by parts would require four separate Easys, which is tedious. $\int x e^x \, dx$ and $\int \ln x \, dx$ require only one Easy of standard integration by parts, while $\int e^x \sin x \, dx$ involves a circular integration pattern that is better handled by the standard method with a trick.

📖 Explanation: To evaluate $\int (x^2 + 3x - 1) \cos(2x) \, dx$, differentiate the polynomial: $f(x) = x^2 + 3x - 1$, f'(x) = 2x + 3, f''(x) = 2, f'''(x) = 0. Integrate $\cos(2x)$ repeatedly: $g(x) = \frac{1}{2}\sin(2x)$, $\int g(x) dx = -\frac{1}{4}\cos(2x)$, $\int \int g(x) dx dx = -\frac{1}{8}\sin(2x)$. Applying the alternating signs $(+,-,+)$: $x^2 \cdot \frac{1}{2}\sin(2x) - (2x+3) \cdot (-\frac{1}{4}\cos(2x)) + 2 \cdot (-\frac{1}{8}\sin(2x))$. The coefficient of $\sin(2x)$ from the first and third terms is $\frac{x^2}{2} - \frac{1}{4}$. However, after simplification, the polynomial coefficient of $\sin(2x)$ is $-\frac{1}{2}x^2 - \frac{3}{2}x + \frac{1}{2}$.

📖 Explanation: The student is incorrect because tabular integration by parts can be extended to handle periodic functions like $e^{3x}$ and $\sin(2x)$. While neither function differentiates to zero, the process creates a cycle of derivatives and integrals that eventually returns to the original product after a few steps. By setting up the table, multiplying with alternating signs, and then solving for the original integral, one can evaluate it efficiently. This technique is often used when the integrand is a product of exponential and trigonometric functions, showing that tabular integration is not limited to polynomial cases.

📖 Explanation: The alternating signs $(+,-,+,-)$ are crucial in tabular integration by parts because they correspond to the pattern $+, -, +, -, \ldots$ from the integration by parts formula. If all signs were positive, the terms would not correctly combine to represent the integral. For $x^2 \sin x$, the correct result with alternating signs is $-x^2 \cos x + 2x \sin x + 2\cos x + C$. With all positive signs, the result would incorrectly be $x^2 \cos x + 2x \sin x - 2\cos x + C$, which does not differentiate back to the original integrand. The sign pattern is derived from the repeated Easy of $\int u dv = uv - \int v du$.

📖 Explanation: A polynomial of degree $n$ has the property that its $(n+1)$-th derivative is zero. For a degree 4 polynomial, the derivatives are: first derivative (degree 3), second (degree 2), third (degree 1), fourth (degree 0), and fifth (zero). Therefore, it takes 5 differentiations to reach zero. This is the fundamental reason why tabular integration by parts terminates for polynomial functions: the polynomial column eventually becomes zero, leaving no further terms to integrate. This property makes tabular integration particularly powerful for integrals involving polynomials.

📖 Explanation: When differentiating $x^3$ repeatedly, the derivatives are: $3x^2, 6x, 6, 0$. When integrating $e^{2x}$ repeatedly, the integrals are: $\frac{1}{2}e^{2x}, \frac{1}{4}e^{2x}, \frac{1}{8}e^{2x}, \frac{1}{16}e^{2x}$. The row before the zero row (third derivative of $x^3$) is the pair $(6, \frac{1}{8}e^{2x})$. This row is multiplied by the sign for that row and contributes to the final integral. The next row would be $(0, \frac{1}{16}e^{2x})$, which contributes nothing (since 0 times anything is 0), effectively terminating the process.

📖 Explanation: Using tabular integration, differentiate $x^4$ to get $4x^3, 12x^2, 24x, 24, 0$. Integrate $e^{-x}$ to get $-e^{-x}, e^{-x}, -e^{-x}, e^{-x}, -e^{-x}$. Apply signs $+, -, +, -, +$ to the products. The result is: $x^4(-e^{-x}) - 4x^3(e^{-x}) + 12x^2(-e^{-x}) - 24x(e^{-x}) + 24(-e^{-x}) = -e^{-x}(x^4 + 4x^3 + 12x^2 + 24x + 24) + C$. The coefficient of $e^{-x}$ is the negative of this polynomial, so the coefficient is $-x^4 - 4x^3 - 12x^2 - 24x - 24$.

📖 Explanation: Tabular integration by parts is most effective when one of the functions in the product is a polynomial that can be differentiated repeatedly to zero. This is its primary limitation and why it is not universally applicable. While it can be extended to handle products of exponentials and trigonometric functions through a cyclic process, the standard, most efficient use of the method requires a polynomial factor. It can handle definite integrals by evaluating the antiderivative at the limits, and it is not inherently more complicated than standard integration by parts for suitable integrands.

📖 Explanation: Applying the signs $+, -, +$ to the products: $(x^2 - 4x + 1) \cdot \frac{1}{3}e^{3x} - (2x - 4) \cdot \frac{1}{9}e^{3x} + 2 \cdot \frac{1}{27}e^{3x}$. Simplifying: $\frac{x^2}{3}e^{3x} - \frac{4x}{3}e^{3x} + \frac{1}{3}e^{3x} - \frac{2x}{9}e^{3x} + \frac{4}{9}e^{3x} + \frac{2}{27}e^{3x}$. Combining like terms: $\frac{x^2}{3}e^{3x} + (-\frac{4}{3} - \frac{2}{9})x e^{3x} + (\frac{1}{3} + \frac{4}{9} + \frac{2}{27})e^{3x} = \frac{x^2}{3}e^{3x} - \frac{14x}{9}e^{3x} + \frac{26}{27}e^{3x} + C$.

📖 Explanation: The correct Easy of tabular integration for $\int x^3 \cos x \, dx$ with alternating signs $+, -, +, -$ yields $x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6\cos x + C$. The student's answer has the signs reversed for the first and third terms, indicating they likely started with a minus sign ($-, +, -, +$). This pattern would result in $-x^3 \sin x - 3x^2 \cos x + 6x \sin x + 6\cos x + C$. The student's result is incorrect because it does not differentiate back to the original integrand, confirming an error in the sign assignment, which is a common mistake in tabular integration.

📖 Explanation: For $\int x^5 e^x \, dx$, tabular integration by parts organizes the repeated differentiations of $x^5$ and integrations of $e^x$ into a neat table, which significantly reduces the chance of algebraic errors that can occur when applying integration by parts five times. While it still requires multiple steps, the tabular format makes the process systematic and easier to track, especially for polynomials of high degree. It doesn't necessarily require fewer steps, but it makes the process more transparent and less prone to mistakes. The antiderivative is not more compact; in fact, it is a polynomial times $e^x$.

📖 Explanation: When integrating products of exponential and trigonometric functions, the trigonometric function $\cos(3x)$ cycles through derivatives and integrals that return to the original function after a few steps. This cyclic nature means the table does not terminate; instead, it creates a loop. By setting up the table and applying the alternating signs, the result will include the original integral on the right side. This allows you to solve for the integral algebraically. The exponential function does not differentiate to zero, and the table doesn't have infinite rows because you stop when the cycle repeats, usually after two or four steps.

📖 Explanation: The LIATE rule suggests choosing $\ln x$ (Logarithmic) as $u$ because its derivative $1/x$ is simpler. Tabular integration can be applied here because $\ln x$ differentiates to a rational function, and while it may not terminate to zero, the process can still be useful. In this case, it requires two steps: differentiate $\ln x$ to $1/x$, then to $-1/x^2$, and integrate $x^2$ to $x^3/3$, then to $x^4/12$. The method still works because the resulting integral $\int x^2 \, dx$ after the first Easy is simpler, but tabular integration is not strictly necessary here; a single Easy of standard integration by parts would suffice.

📖 Explanation: Differentiate $4x^3 - 6x$ to get $12x^2 - 6, 24x, 24, 0$. Integrate $e^{-x}$ to get $-e^{-x}, e^{-x}, -e^{-x}, e^{-x}, -e^{-x}$. Apply signs $+, -, +, -$ to the products: $(4x^3 - 6x)(-e^{-x}) - (12x^2 - 6)(e^{-x}) + (24x)(-e^{-x}) - 24(e^{-x})$. Simplify: $-4x^3 e^{-x} + 6x e^{-x} - 12x^2 e^{-x} + 6e^{-x} - 24x e^{-x} - 24e^{-x}$. Combine like terms: $-e^{-x}(4x^3 + 12x^2 - 18x + 18)$. There appears to be a misalignment in the provided options; the correct simplification leads to $-e^{-x}(4x^3 + 12x^2 - 18x + 18) + C$. Option A is the closest, but it has a different constant term. Let's re-evaluate: The constant term from the second product is $-6e^{-x}$ (from $- (12x^2 - 6)(e^{-x})$ gives $-12x^2 e^{-x} + 6e^{-x}$). The third product is $-24x e^{-x}$. The fourth is $-24e^{-x}$. Sum of constants: $6 - 24 = -18$. So the coefficient of $e^{-x}$ is $-(4x^3 + 12x^2 - 18x + 18)$. None of the options match exactly. The correct option should be $-e^{-x}(4x^3 + 12x^2 - 18x + 18) + C$. This highlights the need for careful simplification in tabular integration.

📖 Explanation: The complexity of the antiderivative from tabular integration is determined by the degree of the polynomial, not its specific roots or the values of its derivatives at specific points. A polynomial of degree 2 will always produce an antiderivative with three terms (from p(x), p'(x), p''(x)), regardless of where its roots are. The roots of the polynomial or its derivatives do not affect the number of terms in the final result, only the values of the coefficients. Having a root at $x = 0$ for $p(x)$ would affect the constant term but not the number of terms. Therefore, all statements about complexity based on roots are incorrect.

📖 Explanation: The number of terms in the antiderivative is determined by the number of non-zero derivatives of $x^n$, which is $n$ derivatives plus the original function, giving $n+1$ terms. For example, for $n=2$, the derivatives are $x^2, 2x, 2, 0$, which produces three non-zero rows, thus three terms in the final antiderivative. This is a direct consequence of the tabular method: each row (except the zero row) contributes a term. The coefficient $a$ in $\sin(ax)$ affects the values of the coefficients but not the number of terms.

📖 Explanation: While tabular integration by parts is a powerful organizational tool that can reduce errors and make repeated Easys more systematic, the assertion that it is 'always faster' is subjective and depends on the individual's familiarity with both methods. For a highly skilled practitioner, standard integration by parts might be just as fast, especially for lower-degree polynomials where the pattern is obvious. Tabular integration is not inherently faster; it primarily reduces the cognitive load by providing a clear structure. The actual speed is a matter of personal proficiency and the specific problem context. Therefore, the claim is false because it makes an absolute statement about speed, which is not guaranteed.

📖 Explanation: Tabular integration by parts is most advantageous when the integrand is a product of a high-degree polynomial and a function that is easily integrated repeatedly, such as $\sin x$ or $\cos x$. A polynomial of degree 5 integrated with $\sin x$ requires multiple Easys of integration by parts, which are efficiently managed by the tabular method. In contrast, an integrand with a logarithmic function (like $\ln x$) is typically integrated with a single Easy of standard integration by parts, and products of exponentials and trigonometric functions are often handled with a cyclic integration technique rather than simple tabular iteration.

📖 Explanation: The correct Easy of signs $+, -, +, -$ for $\int x^3 e^{-x} \, dx$ yields $-e^{-x}(x^3 + 3x^2 + 6x + 6) + C$. If the signs are reversed to $-, +, -, +$, the result becomes the negative of the correct answer, which is $e^{-x}(x^3 + 3x^2 + 6x + 6) + C$. This reversed-sign pattern is a common error. It changes the sign of every term in the antiderivative, making it incorrect because it will not differentiate back to the original integrand. The correct sign pattern is crucial for the validity of the result.

📖 Explanation: To find the sign pattern, look at the terms: The first term is $(x^2)(\frac{1}{3}\sin(3x))$, so the sign for row 0 is $+$. The second term is $(2x)(-\frac{1}{9}\cos(3x))$, which simplifies to $-\frac{2x}{9}\cos(3x)$. However, the given answer has $+\frac{2x}{9}\cos(3x)$. This means the sign for row 1 must be $-$, because $-\frac{1}{9} \cdot (-1) = \frac{1}{9}$. So the sign is $-$. The third term comes from $2 \cdot (-\frac{1}{27}\sin(3x))$. The given answer has $-\frac{2}{27}\sin(3x)$, so the sign for row 2 must be $+$. Therefore, the sign pattern is $+, -, +$.

📖 Explanation: For $\int x^2 e^x \, dx$, the derivatives are $x^2, 2x, 2, 0$, and the integrals of $e^x$ are all $e^x$. The correct sign pattern is $+, -, +$. Therefore, the result is $x^2 e^x - 2x e^x + 2e^x + C$. If the student forgot the signs but still had the correct structure, they might have used all plus signs, leading to $x^2 e^x + 2x e^x + 2e^x + C$, or some other combination. The correct answer is the one with the alternating signs $+, -, +$. Interestingly, the student's result $x^2 e^x - 2x e^x + 2e^x$ actually has the correct signs, so the student did not forget them; they might have made a different mistake. The question states they forgot the signs but produced a result with the signs, creating a paradox. The correct result is $x^2 e^x - 2x e^x + 2e^x + C$, which shows the student's answer is actually correct. The question is designed to test whether the student recognizes the correct sign pattern.

📖 Explanation: When integrating $p(x) e^{kx}$, tabular integration differentiates $p(x)$ (a degree 3 polynomial) to a degree 2 polynomial, then degree 1, then a constant, then zero. Each derivative is multiplied by an integral of $e^{kx}$ (which is always a constant multiple of $e^{kx}$) and summed. Since the derivatives of a degree 3 polynomial are polynomials of degree 3, 2, 1, and 0, the sum of these terms is still a polynomial of degree 3 (because the leading term from the first product does not cancel with anything else). Therefore, the antiderivative is of the form $e^{kx}$ times a polynomial of degree 3.

📖 Explanation: The correct Easy for $\int (x^4 + 1) \sin x \, dx$ involves differentiating $x^4 + 1$ to $4x^3, 12x^2, 24x, 24, 0$, and integrating $\sin x$ to $-\cos x, -\sin x, \cos x, \sin x, -\cos x$. With signs $+, -, +, -, +$, the result is: $(x^4 + 1)(-\cos x) - 4x^3(-\sin x) + 12x^2(\cos x) - 24x(\sin x) + 24(-\cos x)$. Simplifying gives $-(x^4 + 1)\cos x + 4x^3 \sin x + 12x^2 \cos x - 24x \sin x - 24\cos x + C$. The student's result has $-24\cos x$ instead of $-24\cos x$, which is correct. Let's re-check the student's result: $-(x^4 + 1)\cos x + 4x^3 \sin x + 12x^2 \cos x - 24x \sin x - 24\cos x + C$. This matches the correct result. The student made no error. The question is flawed in stating there is an error.

📖 Explanation: When integrating a polynomial times $\sin(2x)$, the tabular method produces a result that contains both sine and cosine terms. This is because the derivatives of the polynomial produce constant terms, and the integrals of $\sin(2x)$ alternate between $-\frac{1}{2}\cos(2x)$ and $-\frac{1}{4}\sin(2x)$, creating a mix. The final answer will have terms with $\sin(2x)$ from some rows and $\cos(2x)$ from other rows, making it a combination of both trigonometric functions. This is a general property of such integrals.

📖 Explanation: Using the antiderivative $F(x) = -x^2 \cos x + 2x \sin x + 2\cos x$, evaluate from 0 to $\pi$: $F(\pi) - F(0) = (-\pi^2 \cos \pi + 2\pi \sin \pi + 2\cos \pi) - (0 + 0 + 2\cos 0) = (-\pi^2(-1) + 0 - 2) - (2) = \pi^2 - 2 - 2 = \pi^2 - 4$. This demonstrates how tabular integration can be directly applied to definite integrals. The process of finding the antiderivative is the same; the limits are applied afterward.

📖 Explanation: For $\int e^x \sin x \, dx$, the tabular method differentiates $e^x$ (which cycles) and integrates $\sin x$ (which also cycles). Setting up the table and applying the signs yields: $e^x(-\cos x) - e^x(-\sin x) +$ (the original integral). This leads to $\int e^x \sin x \, dx = -e^x \cos x + e^x \sin x - \int e^x \sin x \, dx$. Solving for the integral gives $2\int e^x \sin x \, dx = e^x(\sin x - \cos x)$, so $\int e^x \sin x \, dx = \frac{e^x(\sin x - \cos x)}{2} + C$. This method is an extension of tabular integration to non-terminating functions.

📖 Explanation: Tabular integration by parts is derived directly from the standard integration by parts formula $\int u dv = uv - \int v du$. Understanding this derivation is essential for correctly applying the method, especially when determining the sign pattern, identifying when to stop the table, and extending the method to non-polynomial functions. Viewing it as a mere shortcut without understanding its mathematical foundation can lead to errors in complex problems. The method is a structured Easy of the product rule in integral form, not a separate or independent technique.

📖 Explanation: If a polynomial differentiates to zero in 4 steps, this means the polynomial has degree 3 (since the 4th derivative is zero). The table includes the original polynomial as the first row, then each derivative as a subsequent row, ending with the zero row. Therefore, the number of rows is 4 (the non-zero derivatives) + 1 (the zero row) = 5 rows. Alternatively, if 'differentiates to zero in 4 steps' means the 4th derivative is zero, then there are 4 non-zero rows (the function and its first 3 derivatives) and 1 zero row, totaling 5 rows.

📖 Explanation: While $\int x^3 \cos x \, dx$ and $\int x^2 \ln x \, dx$ are straightforward Easys of tabular integration (with a polynomial factor), $\int e^{2x} \sin(3x) \, dx$ is Hard because neither function differentiates to zero. The tabular method must be extended to handle the cyclic nature of the derivatives and integrals of the exponential and trigonometric functions. This requires setting up the table, identifying the cycle, and solving an algebraic equation for the original integral. $\int \cos x \sin x \, dx$ is a simple substitution problem and not a typical candidate for tabular integration. Thus, the cyclic case is the most complex Easy of the method.

📖 Explanation: For $\int x^3 \cos x \, dx$, differentiate $x^3$ to get $3x^2, 6x, 6, 0$. Integrate $\cos x$ to get $\sin x, -\cos x, -\sin x, \cos x$. Applying the signs $+, -, +, -$, the products are: $x^3 \cdot \sin x$ (row 0, sign +), $3x^2 \cdot (-\cos x)$ (row 1, sign -), $6x \cdot (-\sin x)$ (row 2, sign +), $6 \cdot (\cos x)$ (row 3, sign -). So the expression is $x^3 \sin x - 3x^2(-\cos x) + 6x(-\sin x) - 6(\cos x)$, which simplifies to $x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6\cos x + C$.

📖 Explanation: Let's trace the correct process: Differentiate $x^2 + 2x$ to $2x + 2, 2, 0$. Integrate $\cos(2x)$ to $\frac{1}{2}\sin(2x), -\frac{1}{4}\cos(2x), -\frac{1}{8}\sin(2x)$. Applying signs $+, -, +$, the correct result is $(x^2 + 2x)\frac{1}{2}\sin(2x) - (2x+2)(-\frac{1}{4}\cos(2x)) + 2(-\frac{1}{8}\sin(2x))$, which simplifies to $\frac{x^2}{2}\sin(2x) + x\sin(2x) + \frac{x}{2}\cos(2x) + \frac{1}{2}\cos(2x) - \frac{1}{4}\sin(2x) + C$. The student's answer has $\frac{3x}{2}\cos(2x)$ and $-\frac{3}{4}\sin(2x)$, which suggests they incorrectly combined terms or used the wrong integral values for $\cos(2x)$. The coefficient of $\cos(2x)$ should be $\frac{x}{2} + \frac{1}{2}$, not $\frac{3x}{2}$. This indicates an error in the integration of $\cos(2x)$ or the signs.

📖 Explanation: The alternating signs in tabular integration by parts come from the repeated Easy of the integration by parts formula $\int u dv = uv - \int v du$. If $u = p(x)$ and $dv = q(x) dx$, the sign for the $k$-th term is $(-1)^k$, where $k$ is the order of the derivative. This is because each Easy of the formula introduces a minus sign. Therefore, the first term (k=0) is positive, the second (k=1) is negative, the third (k=2) is positive, and so on. This alternation is a fundamental property of the method, derived directly from the product rule.