Quiz:Integration by parts

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For background, see integration by parts.

Statement

Formal manipulation version for definite integration in function notation

1 Suppose F,G are continuously differentiable functions defined on all of \R. Suppose a,b are real numbers with a < b. Suppose, further, that G(x) is identically zero everywhere except on the open interval (a,b). Then, what can we say about the relationship between the numbers P = \int_a^b F(x)G'(x) \,dx and Q = \int_a^b F'(x)G(x) \, dx?

P = Q
P = -Q
PQ = 0
P = 1 - Q
PQ = 1

Key observations

Equivalence of integration problems

1 Which of the following is not true (the ones that are true can be deduced from integration by parts)?
RELATED COMPUTATIONAL QUESTIONS (well, sort of): \int (e^x)(1/x^2) \, dx, \int x \tan x \, dx

We can compute an expression for the antiderivative of the pointwise product of functions fg based on knowledge of expressions for f, g, and their antiderivatives.
Suppose F and G are everywhere differentiable. Given an expression for the antiderivative for the pointwise product of functions \! F'G, we can obtain an expression for the antiderivative for the pointwise product \! FG'.
If F is a one-to-one function, we can find an antiderivative for F^{-1} in terms of F^{-1} and an antiderivative for F.

For more quiz questions on the theme of equivalence of integration problems, see Quiz:Equivalence of integration problems.

Repeated use of integration by parts and the circular trap

1 Suppose we start with a product of two functions F and g. Apply integration by parts k times. Start off by taking F as the part to differentiate and g 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^{th} derivative of F and the k^{th} derivative of g.
It is a product of the k^{th} derivative of F and the k^{th} antiderivative of g.
It is a product of the k^{th} antiderivative of F and the k^{th} derivative of g.
It is a product of the k^{th} antiderivative of F and the k^{th} antiderivative of g.

2 Consider the integration \int p(x) q''(x) \, dx. Apply integration by parts twice, first taking p as the part to differentiate, and q as the part to integrate, and then again apply integration by parts to avoid a circular trap. What can we conclude?
RELATED COMPUTATIONAL QUESTIONS: \int x e^x \, dx (think p(x) = x, q(x) = e^x), \int x^2 (\sin x) \, dx (think p(x) = x^2, q(x) = -\sin x)

\int p(x) q''(x) \, dx = \int p''(x) q(x) \, dx
\int p(x) q''(x) \, dx = \int p'(x) q'(x) \, dx - \int p''(x) q(x) \, dx
\int p(x)q''(x) \,dx = p'(x)q'(x) - \int p''(x) q(x)\, dx
\int p(x)q''(x) \,dx = p(x)q'(x) - p'(x)q(x) + \int p''(x) q(x)\, dx
\int p(x)q''(x) \,dx = p(x)q'(x) - p'(x)q(x) - \int p''(x) q(x)\, dx

3 Which of the following is an incorrect way of applying integration by parts twice?

After applying integration by parts once, we get a new product. Choose as the part to integrate the factor in the product arising from integration, and as the part to differentiate the factor in the product arising from differentiation.
After applying integration by parts once, we get a new product. Choose as the part to differentiate the factor in the product arising from integration, and as the part to integrate the factor in the product arising from differentiation.
Neither method is incorrect in general. The first method is used for straightforward integrations and the second method is used for the recursive version of integration by parts.

Recursive version of integration by parts

See the questions in the next section, #Choosing the parts to integrate and differentiate.

Integration by parts is not the exclusive strategy for products

1 Which of the following correctly describes the relationship between integration by parts and integration by u-substitution when deciding to integrate a pointwise product of functions or a composite of two functions?

Integration by parts, which is obtained from the product rule for differentiation, is the exclusive strategy for integrating products. Integration by u-substitution, which is obtained from the chain rule for differentiation, is the exclusive strategy for integrating composites.
Integration by parts, which is obtained from the product rule for differentiation, is the exclusive strategy for integrating composites. Integration by u-substitution, which is obtained from the chain rule for differentiation, is the exclusive strategy for integrating products.
Both methods are useful for both types of integrations. Specifically, integration by parts helps with certain kinds of products and composites, and integration by u-substitution helps with certain kinds of products.

2 Which of the following functions of x can be integrated with respect to x without integration by parts, and purely using integration by u-substitution and the knowledge of the antiderivative of the cosine function?

x^2 \cos x
x \cos^2 x
x \cos (x^2)
x^2 \cos (x^2)
x^2 \cos^2 x

Choosing the parts to integrate and differentiate

Products and typical strategies

Review the strategies for products and determining the parts to integrate and differentiate: [SHOW MORE]

1 Suppose p is a polynomial function. In order to find the indefinite integral for a function of the form x \mapsto 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?
RELATED COMPUTATIONAL PRACTICE: \int x^2 \sin x \,dx, \int x \sin x \, dx, \int (x^2 + 1)^2 \cos x \, dx, \int x \cos^2 x \, dx, \int 2x \cos (3x) \, dx

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

2 Suppose p is a polynomial function. In order to find the indefinite integral for a function of the form x \mapsto p(x)\exp(x), the general strategy, which always works, is to take p(x) as the part to differentiate and \exp(x) as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works? RELATED COMPUTATIONAL PRACTICE: \int x e^x \, dx, \int x^2 e^x\, dx, \int (x +3) e^{-4x} \, dx

\exp can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way.
\exp can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero.
\exp and polynomials can both be repeatedly differentiated.
\exp and polynomials can both be repeatedly integrated.

3 Consider the function x \mapsto \exp(x) \sin x. This function can be integrated using integration by parts. What can we say about how integration by parts works?
RELATED COMPUTATIONAL PRACTICE: \int e^{-x} \cos x \, dx, \int e^x \cos x \, dx, \int e^x \sin^2 x \, dx

We choose \exp as the part to integrate and \sin as the part to differentiate, and apply this process once to get the answer directly.
We choose \exp as the part to integrate and \sin 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 \exp as the part to integrate and \sin as the part to differentiate, and apply this process twice to get the answer directly.
We choose \exp as the part to integrate and \sin 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.

4 Suppose p is a polynomial function. Consider the function x \mapsto p(x) \ln x for x > 0. This function can be integrated using integration by parts and knowledge of how to integrate polynomials. Which of the following is the best integration strategy?
RELATED COMPUTATIONAL PRACTICE: \int \ln x \, dx, \int x \ln x \, dx, \int (x^2 - 2)\ln(x + 3) \,dx.

Take p as the part to differentiate and \ln as the part to integrate. Integration by parts needs to be applied just once.
Take p as the part to integrate and \ln as the part to differentiate. Integration by parts needs to be applied just once.
Take p as the part to differentiate and \ln as the part to integrate. Integration by parts needs to be applied as many times as the degree of p.
Take p as the part to integrate and \ln as the part to differentiate. Integration by parts needs to be applied as many times as the degree of p.

5 Suppose p and q are polynomial functions. Consider the function x \mapsto p(x) q(\ln x) for x > 0. This function can be 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?
RELATED COMPUTATIONAL PRACTICE: \int x^3(\ln x)^2 \, dx, \int (x^2 + 1)((\ln x)^2 + 1)) \, dx

The number of times equals the sum of degrees of p and q.
The number of times equals the product of degrees of p and q.
The number of times equals the degree of p.
The number of times equals the degree of q.

6 Suppose p and q are polynomial functions. Consider the function x \mapsto p(x) \ln(q(x)). Assume we are integrating over a domain where q(x) is positive. How can we integrate this function, and why does the method work?
RELATED COMPUTATIONAL PRACTICE: \int \ln(x^2 + 1) \, dx, \int x \ln(x^3 + 1) \, dx

We can integrate the function by applying integration by parts repeatedly, taking \ln(q(x)) as the part to differentiate and p(x) as the part to integrate. The number of times we need to apply integration by parts is the degree of q.
We can integrate the function by applying integration by parts, taking \ln(q(x)) as the part to differentiate and p(x) as the part to integrate, and then integrating the rational function so obtained. We use the fact that there is a strategy for integrating any rational function.
We can integrate the function by applying integration by parts repeatedly, taking \ln(q(x)) as the part to integrate and p(x) as the part to differentiate. The number of times we need to apply integration by parts is the degree of p.
We can integrate the function by applying integration by parts, taking \ln(q(x)) as the part to integrate and p(x) as the part to differentiate, and then integrating the rational function so obtained. We use the fact that there is a strategy for integrating any rational function.