Quiz:Integration by parts - Revision history

Vipul at 02:22, 21 November 2012

2012-11-21T02:22:41Z

← Older revision		Revision as of 02:22, 21 November 2012
Line 4:		Line 4:
	===Formal manipulation version for definite integration in function notation===		===Formal manipulation version for definite integration in function notation===
	<quiz display=simple>		<quiz display=simple>
	{Suppose <math>F,G</math> are ~~continuous~~ functions defined on all of <math>\R</matH>. Suppose <math>a,b</math> are real numbers with <math>a < b</math>. Suppose, further, that <math>G(x)</math> is identically zero everywhere except on the open interval <math>(a,b)</math>. Then, what can we say about the relationship between the numbers <math>P = \int_a^b F(x)G'(x) \,dx</math> and <math>Q = \int_a^b F'(x)G(x) \, dx</math>?		{Suppose <math>F,G</math> are continuously differentiable functions defined on all of <math>\R</matH>. Suppose <math>a,b</math> are real numbers with <math>a < b</math>. Suppose, further, that <math>G(x)</math> is identically zero everywhere except on the open interval <math>(a,b)</math>. Then, what can we say about the relationship between the numbers <math>P = \int_a^b F(x)G'(x) \,dx</math> and <math>Q = \int_a^b F'(x)G(x) \, dx</math>?
	\|type="()"}		\|type="()"}
	- <math>P = Q</math>		- <math>P = Q</math>

Vipul: /* Products and typical strategies */

2012-02-20T04:59:07Z

Products and typical strategies

← Older revision		Revision as of 04:59, 20 February 2012
Line 111:		Line 111:
	- We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> as the part to differentiate, and apply this process once to get the answer directly.		- We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> as the part to differentiate, and apply this process once to get the answer directly.
	- We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> 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 <matH>\exp</math> as the part to integrate and <math>\sin</math> 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 <matH>\exp</math> as the part to integrate and <math>\sin</math> as the part to differentiate, and apply this process ~~twicce~~ to get the answer directly.		- We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> as the part to differentiate, and apply this process twice to get the answer directly.
	+ We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> 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.		+ We choose <matH>\exp</math> as the part to integrate and <math>\sin</math> 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.
	\|\| Applying integration by parts twice, we get <math>\int e^x \sin x \,dx = e^x (\sin x - \cos x) - \int e^x\sin x \, dx</math>. Now, rearrange and simplify.		\|\| Applying integration by parts twice, we get <math>\int e^x \sin x \,dx = e^x (\sin x - \cos x) - \int e^x\sin x \, dx</math>. Now, rearrange and simplify.

Vipul: /* Products and typical strategies */

2012-02-20T04:57:29Z

Products and typical strategies

← Older revision		Revision as of 04:57, 20 February 2012
Line 119:		Line 119:
	\|type="()"}		\|type="()"}
	- Take <math>p</math> as the part to differentiate and <math>\ln</math> as the part to integrate. Integration by parts needs to be applied just once.		- Take <math>p</math> as the part to differentiate and <math>\ln</math> as the part to integrate. Integration by parts needs to be applied just once.
	+ Take <math>p</math> as the part to integrate and <math>\ln</math> as the part to ~~integrate~~. Integration by parts needs to be applied just once.		+ Take <math>p</math> as the part to integrate and <math>\ln</math> as the part to differentiate. Integration by parts needs to be applied just once.
	\|\| Let <math>P</math> be a polynomial antiderivative for <math>p</math> chosen to have no constant term. Then, the first application of integration by parts gives: <math>(\ln x)P(x) - \int \frac{P(x)}{x} \, dx</math>. Since <math>P</math> has no constant term, <math>P(x)/x</math> is also a polynomial, and can be integrated by the usual method of integrating polynomials. Note that we needed to apply integration by parts only once.		\|\| Let <math>P</math> be a polynomial antiderivative for <math>p</math> chosen to have no constant term. Then, the first application of integration by parts gives: <math>(\ln x)P(x) - \int \frac{P(x)}{x} \, dx</math>. Since <math>P</math> has no constant term, <math>P(x)/x</math> is also a polynomial, and can be integrated by the usual method of integrating polynomials. Note that we needed to apply integration by parts only once.
	- Take <math>p</math> as the part to differentiate and <math>\ln</math> as the part to integrate. Integration by parts needs to be applied as many times as the degree of <math>p</math>.		- Take <math>p</math> as the part to differentiate and <math>\ln</math> as the part to integrate. Integration by parts needs to be applied as many times as the degree of <math>p</math>.
	- Take <math>p</math> as the part to integrate and <math>\ln</math> as the part to ~~integrate~~. Integration by parts needs to be applied as many times as the degree of <math>p</math>.		- Take <math>p</math> as the part to integrate and <math>\ln</math> as the part to differentiate. Integration by parts needs to be applied as many times as the degree of <math>p</math>.

	{Suppose <math>p</math> and <math>q</math> are polynomial functions. Consider the function <math>x \mapsto p(x) q(\ln x)</math> for <math>x > 0</math>. 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?<br>		{Suppose <math>p</math> and <math>q</math> are polynomial functions. Consider the function <math>x \mapsto p(x) q(\ln x)</math> for <math>x > 0</math>. 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?<br>

Vipul: /* Equivalence of integration problems */

2012-02-20T04:53:07Z

Equivalence of integration problems

← Older revision		Revision as of 04:53, 20 February 2012
Line 19:		Line 19:

	<quiz display=simple>		<quiz display=simple>
	{Which of the following is ''not'' true (the ones that are true can be deduced from integration by parts)?		{Which of the following is ''not'' true (the ones that are true can be deduced from integration by parts)?<br>'''RELATED COMPUTATIONAL QUESTIONS''' (well, sort of): <math>\int (e^x)(1/x^2) \, dx</math>, <math>\int x \tan x \, dx</math>
	\|type="()"}		\|type="()"}
	+ We can compute an expression for the antiderivative of the [[pointwise product of functions]] <math>fg</math> based on knowledge of expressions for <math>f</math>, <math>g</math>, and their antiderivatives.		+ We can compute an expression for the antiderivative of the [[pointwise product of functions]] <math>fg</math> based on knowledge of expressions for <math>f</math>, <math>g</math>, and their antiderivatives.

Vipul: /* Repeated use of integration by parts and the circular trap */

2012-02-20T04:49:56Z

Repeated use of integration by parts and the circular trap

← Older revision		Revision as of 04:49, 20 February 2012
Line 31:		Line 31:

	<quiz display=simple>		<quiz display=simple>
	{Suppose we start with a product of two functions <math>F</math> and <math>g</math>. Apply integration by parts <math>k</math> times. Start off by taking <math>F</math> as the part to differentiate and <math>g</math> 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 ~~antiderivatve~~ that don't involve integral signs.		{Suppose we start with a product of two functions <math>F</math> and <math>g</math>. Apply integration by parts <math>k</math> times. Start off by taking <math>F</math> as the part to differentiate and <math>g</math> 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.
	\|type="()"}		\|type="()"}
	- It is a product of the <math>k^{th}</math> derivative of <math>F</math> and the <math>k^{th}</math> derivative of <math>g</math>.		- It is a product of the <math>k^{th}</math> derivative of <math>F</math> and the <math>k^{th}</math> derivative of <math>g</math>.
Line 38:		Line 38:
	- It is a product of the <math>k^{th}</math> antiderivative of <math>F</math> and the <math>k^{th}</math> antiderivative of <math>g</math>.		- It is a product of the <math>k^{th}</math> antiderivative of <math>F</math> and the <math>k^{th}</math> antiderivative of <math>g</math>.

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

Vipul: /* Formal manipulation version for definite integration in function notation */

2012-02-20T04:46:15Z

Formal manipulation version for definite integration in function notation

← Older revision		Revision as of 04:46, 20 February 2012
Line 5:		Line 5:
	<quiz display=simple>		<quiz display=simple>
	{Suppose <math>F,G</math> are continuous functions defined on all of <math>\R</matH>. Suppose <math>a,b</math> are real numbers with <math>a < b</math>. Suppose, further, that <math>G(x)</math> is identically zero everywhere except on the open interval <math>(a,b)</math>. Then, what can we say about the relationship between the numbers <math>P = \int_a^b F(x)G'(x) \,dx</math> and <math>Q = \int_a^b F'(x)G(x) \, dx</math>?		{Suppose <math>F,G</math> are continuous functions defined on all of <math>\R</matH>. Suppose <math>a,b</math> are real numbers with <math>a < b</math>. Suppose, further, that <math>G(x)</math> is identically zero everywhere except on the open interval <math>(a,b)</math>. Then, what can we say about the relationship between the numbers <math>P = \int_a^b F(x)G'(x) \,dx</math> and <math>Q = \int_a^b F'(x)G(x) \, dx</math>?
			\|type="()"}
	- <math>P = Q</math>		- <math>P = Q</math>
	+ <math>P = -Q</math>		+ <math>P = -Q</math>
Line 12:		Line 13:
	- <math>PQ = 1</math>		- <math>PQ = 1</math>

			</quiz>
	==Key observations==		==Key observations==

Vipul: /* Statement */

2012-02-20T04:45:50Z

Statement

← Older revision		Revision as of 04:45, 20 February 2012
Line 2:		Line 2:

	==Statement==		==Statement==
			===Formal manipulation version for definite integration in function notation===
	No quiz ~~questions~~ on the ~~statement yet~~.		<quiz display=simple>
			{Suppose <math>F,G</math> are continuous functions defined on all of <math>\R</matH>. Suppose <math>a,b</math> are real numbers with <math>a < b</math>. Suppose, further, that <math>G(x)</math> is identically zero everywhere except on the open interval <math>(a,b)</math>. Then, what can we say about the relationship between the numbers <math>P = \int_a^b F(x)G'(x) \,dx</math> and <math>Q = \int_a^b F'(x)G(x) \, dx</math>?
			- <math>P = Q</math>
			+ <math>P = -Q</math>
			\|\| Using integration by parts, we get that <math>P = [F(x)G(x)]_a^b - Q</math>. By assumption, <math>G(x) = 0</math> outside the open interval <math>(a,b)</math>, so <math>G(a) = G(b) = 0</math>. Thus, the expression <math>[F(x)G(x)]_a^b</math> evaluates to zero. We are left with <math>P = -Q</math>.
			- <math>PQ = 0</math>
			- <math>P = 1 - Q</math>
			- <math>PQ = 1</math>

	==Key observations==		==Key observations==

Vipul: /* Repeated use of integration by parts and the circular trap */

2012-02-20T04:38:35Z

Repeated use of integration by parts and the circular trap

← Older revision		Revision as of 04:38, 20 February 2012
Line 30:		Line 30:

	{Consider the integration <math>\int p(x) q''(x) \, dx</math>. Apply integration by parts twice, first taking <math>p</math> as the part to differentiate, and <math>q</math> as the part to integrate, and then again apply integration by parts to avoid a circular trap. What can we conclude?		{Consider the integration <math>\int p(x) q''(x) \, dx</math>. Apply integration by parts twice, first taking <math>p</math> as the part to differentiate, and <math>q</math> as the part to integrate, and then again apply integration by parts to avoid a circular trap. What can we conclude?
			\|type="()"}
	- <math>\int p(x) q''(x) \, dx = \int p''(x) q(x) \, dx</math>		- <math>\int p(x) q''(x) \, dx = \int p''(x) q(x) \, dx</math>
	- <math>\int p(x) q''(x) \, dx = \int p'(x) q'(x) \, dx - \int p''(x) q(x) \, dx</math>		- <math>\int p(x) q''(x) \, dx = \int p'(x) q'(x) \, dx - \int p''(x) q(x) \, dx</math>

Vipul: /* Repeated use of integration by parts and the circular trap */

2012-02-20T04:38:17Z

Repeated use of integration by parts and the circular trap

@@ Line 22: / Line 22: @@
 <quiz display=simple>
 {Which of the following is an ''incorrect'' way of applying integration by parts twice?
 |type="()"}
@@ Line 31: / Line 45: @@
 </quiz>
 ===Recursive version of integration by parts===

Vipul: /* Integration by parts is not the exclusive strategy for products */

2012-02-20T04:30:05Z

Integration by parts is not the exclusive strategy for products

← Older revision		Revision as of 04:30, 20 February 2012
Line 45:		Line 45:
	\|\| For instance, for products of the form <math>h(g(x))g'(x)</math> it is useful to integrate by the u-substitution <math>u = g(x)</math>. On the other hand, for function such as <math>x \sin x</math>, we use integration by parts. Even for pure composites, we may use integration by parts, either directly or combined with integration by u-substitution. For instance, <math>\cos (\ln x)</math>		\|\| For instance, for products of the form <math>h(g(x))g'(x)</math> it is useful to integrate by the u-substitution <math>u = g(x)</math>. On the other hand, for function such as <math>x \sin x</math>, we use integration by parts. Even for pure composites, we may use integration by parts, either directly or combined with integration by u-substitution. For instance, <math>\cos (\ln x)</math>

	{Which of the following functions can be integrated ''without'' integration by parts, and purely using [[integration by u-substitution]] and the knowledge of the antiderivative of the cosine function?		{Which of the following functions of <math>x</math> can be integrated with respect to <math>x</math> ''without'' integration by parts, and purely using [[integration by u-substitution]] and the knowledge of the antiderivative of the cosine function?
	\|type="()"}		\|type="()"}
	- <math>x^2 \cos x</math>		- <math>x^2 \cos x</math>