31. Explain in your own words how the picture proves the theorem. Which notation in theproof is unfamiliar? What do you think it means?Theorem 7 (Integration by Parts (IBP)) If u = u(x) and v = v(x) are functionsof x that have continuous derivatives, thenu dv = uv -v du,where dv = v'(x) dx and du = u'(x) dx.It is important to select the functions u and dv carefully in the setup. Choose thefunction u to be a function that will simplify by calculating the derivative. Choose thefunction dv to be a function that will simplify by calculating the antiderivative.For example, in the integral /xe" dx, put u(x) = x and dv = e*dx. Then calculatedu = u'(x)dx = ldx and take an antiderivative v = ƒ dv = e² (letting the arbitraryconstant C = 0). Apply Theorem 7 to obtainxe" dx =u dv = uv -v du = xe" –e* dx = xe² – e* + C.Sometimes IBP needs to be applied more than once. When applying IBP a secondtime, make sure to make the same type of choices for u and vThe methods of IBP is also useful when integrating inverse transcendental functions,such as the logarithm and the arcsine. If you need to integrate an inverse transcen-dental function, try putting u equal to the inverse transcendental function and takinga derivative to obtain an algebraic function.32.x²e3* dx33.x² cos x d.x34.x³ In 3x dx35.e5z cos(2x) dx36.| In(sin x) dæ1037.In(4 + x²) dx38.a sec?(2r) dr

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Description: 31. Explain in your own words how the picture proves the theorem. Which notation in theproof is unfamiliar? What do you think it means?Theorem 7 (Integration by Parts (IBP)) If u = u(x) and v = v(x) are functionsof x that have continuous derivatives

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32. e* dx

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Topic Video

Question

31. Explain in your own words how the picture proves the theorem. Which notation in the
proof is unfamiliar? What do you think it means?
Theorem 7 (Integration by Parts (IBP)) If u = u(x) and v = v(x) are functions
of x that have continuous derivatives, then
u dv = uv -
v du,
where dv = v'(x) dx and du = u'(x) dx.
It is important to select the functions u and dv carefully in the setup. Choose the
function u to be a function that will simplify by calculating the derivative. Choose the
function dv to be a function that will simplify by calculating the antiderivative.
For example, in the integral /
xe" dx, put u(x) = x and dv = e*dx. Then calculate
du = u'(x)dx = ldx and take an antiderivative v = ƒ dv = e² (letting the arbitrary
constant C = 0). Apply Theorem 7 to obtain
xe" dx =
u dv = uv -
v du = xe" –
e* dx = xe² – e* + C.
Sometimes IBP needs to be applied more than once. When applying IBP a second
time, make sure to make the same type of choices for u and v
The methods of IBP is also useful when integrating inverse transcendental functions,
such as the logarithm and the arcsine. If you need to integrate an inverse transcen-
dental function, try putting u equal to the inverse transcendental function and taking
a derivative to obtain an algebraic function.
32.
x²e3* dx
33.
x² cos x d.x
34.
x³ In 3x dx
35.
e5z cos(2x) dx
36.
| In(sin x) dæ
10
37.
In(4 + x²) dx
38.
a sec?(2r) dr

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