Mathematics for Management -- Supplementary Electronic Materials

Quiz: Repeated Application of Integration by Parts

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1. If you are asked to evaluate the integral \(\int \, x^2 {\rm{e}}^{x} \, \textrm{d} x\), a good first step would be
I) Substitute \(u = {\rm{e}}^{x}\) and \(\textrm{d} u = {\rm{e}}^{x} \textrm{d} x\).
II) Substitute \(u = x^2\) and \(\textrm{d} u = 2x \textrm{d} x\).
III) Integration by parts, with \(u = x^2\) and \(\textrm{d} v = {\rm{e}}^{x} \textrm{d} x\)

Only I) is a good first step Only II) is a good first step Only III) is a good first step A combination of I), II), III) is the best

2. Determine \(\int \, x^2 {\rm{e}}^{x} \, \textrm{d} x\).

\( (x^2-2x+2){\rm{e}}^{-x} + C \) \( (x^2-2x-2){\rm{e}}^{-x} + C \) \( (x^2+2x){\rm{e}}^{-x} + C \) \( (x-1)^2{\rm{e}}^{-x} + C \)

3. How many applications of integration by parts are required to evaluate the integral \(\int \, {\rm{e}}^{4x} \cos(3x) \, \textrm{d} x\)?

\( 4 \) \( 1 \) \( 2 \) \( 3 \)

4. Determine the value of \(\int_0^{\pi/2} \, {\rm{e}}^x \cos(x) \, \textrm{d} x\).

\( \tfrac{1}{2} {\rm{e}}^{\pi/2} - \frac{1}{2} \) \( \tfrac{1}{2} {\rm{e}}^{\pi/2} + \frac{1}{2} \) \( -\tfrac{1}{2} {\rm{e}}^{\pi/2} - \frac{1}{2} \) \( -\tfrac{1}{2} {\rm{e}}^{\pi/2} + \frac{1}{2} \)

5. Determine \(\int \, {\rm{e}}^x \left( 1 + \ln(x) + x \ln(x) \right) \textrm{d} x\).

\( x {\rm{e}}^{x} \ln(x) + C \) \( x^2 {\rm{e}}^x \ln(x) + C \) \( x + {\rm{e}}^x \ln(x) + C \) \( x {\rm{e}}^x + \ln(x) + C \)

6. Suppose we start with a product of two functions \(F(x)\) and \(g(x)\). Apply integration by parts \(k\) times. Start off by taking \(F(x)\) as the part to differentiate and \(g(x)\) as the part to integrate. Each time, take the part to differentiate as the function obtained by differentiation, and the part to integrate as the function obtained by integration. Assuming that the process of repeatedly finding antiderivatives works without a hitch, what can we conclude about the final integrand? Ignore the other times in the antiderivative that don't involve integral signs.

It is a product of the \(k^{\text{th}}\) derivative of \(F(x)\) and the \(k^{\text{th}}\) derivative of \(g(x)\). It is a product of the \(k^{\text{th}}\) derivative of \(F(x)\) and the \(k^{\text{th}}\) antiderivative of \(g(x)\). It is a product of the \(k^{\text{th}}\) anitderivative of \(F(x)\) and the \(k^{\text{th}}\) derivative of \(g(x)\). It is a product of the \(k^{\text{th}}\) antiderivative of \(F(x)\) and the \(k^{\text{th}}\) antiderivative of \(g(x)\).

7. Consider the integration \(\int \, p(x) q''(x) \, \textrm{d} x\). Apply integration by parts, first taking \(p(x)\) as the part to differentiate, and \(q(x)\) as the part to integrate, and the again apply integration by parts to a avoid a circular trap. What can we conclude?

\( \int \, p(x) q''(x) \, \textrm{d} x = \int \, p''(x) q(x) \, \textrm{d} x \) \( \int \, p(x) q''(x) \, \textrm{d} x = \int \, p'(x) q'(x) \, \textrm{d} x - \int \, p''(x) q(x) \, \textrm{d} x \) \( \int \, p(x) q''(x) \, \textrm{d} x = p'(x) q'(x) - \int \, p''(x) q(x) \, \textrm{d} x \) \( \int \, p(x) q''(x) \, \textrm{d} x = p(x) q'(x) - p'(x) q(x)1 + \int \, p''(x) q(x) \, \textrm{d} x \)

8. Suppose \(p(x)\) is a polynomial. In order to find the indefinite integral for a function of the form \(p(x) \sin(x)\), the general strategy, which always works, is to take \(p(x)\) as the part to differentiate and \(\sin(x)\) as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works?

\(\sin(x)\) can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way. \(\sin(x)\) can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero. \(\sin(x)\) and polynomials can both be repeatedly differentiated. \(\sin(x)\)and polynomials can both be repeatedly integrated.

9. Consider the function \({\rm{e}}^x \sin(x)\). This function can be integrated using integration by parts. What can we say about how integration by parts works?

We choose \({\rm{e}}^x\) as the part to integrate and \(\sin(x)\) as the part to differentiate, and apply this process once to get the answer directly. We choose \({\rm{e}}^x\) as the part to integrate and \(\sin(x)\) as the part to differentiate, and apply this process once, then use a recursive method (identify the integrals on the left and right side) to get the answer. We choose \({\rm{e}}^x\) as the part to integrate and \(\sin(x)\) as the part to differentiate, and apply this process twice to get the answer directly. We choose \({\rm{e}}^x\) as the part to integrate and \(\sin(x)\) as the part to differentiate, and apply this process twice, then use a recursive method (identify the integrals on the left and right side) to get the answer.

10. Suppose \( p(x)\) and \( q(x)\) are polynomials. Consider the function \( p(x) q(\ln(x))\) for \( x > 0\). This function can be in integrated using integration by parts and knowledge of how to integrate polynomials. Using the best strategy, how many times do we need to apply integration by parts?

The number of times equals the sum of degrees of \( p(x)\) and \( q(x)\). The number of times equals the product of degrees of \( p(x)\) and \(q(x)\). The number of times equals the degree of \(p(x)\). The number of times equals the degree of \(q(x)\).

Your grade is: __ of 10

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Solution: 1c; 2a; 3c; 4a; 5a; 6b; 7d; 8b; 9d; 10d

Here, a, b, c, d indicate the 1st, 2nd, 3rd, and 4th answer choice, respectively, for the numbered questions.