Calc1131W11T1 - Integration by parts

Sunday, 20 May 2018

7:52 PM

Untitled picture.png Machine generated alternative text:
The integration by parts formula is one of the most useful in the integral calculus. Ifu and t' are both differentiable functions of of then the 
product rule 
O 
tells us that 
du 
t" dc 
If we integrate both sides with respect to x, we get 
du 
We can rearrange this to get 
dv 
u dc 
which is sometimes more compactly written as 
dc + 
uv — 
u dc 
dv 
u dc 
du 
v dc 
t' du. 
Going from the left hand side to the right hand side; we have replaced one integration by another This involves taking the integrand on the left and writing it as the product offtvo functions, where 
dv 
one is a derivative of a function we know. This is useful if the product 
is easier to integrate than u 
The integration by parts formula also works for definite integrals, where the formula is [select all that apply]: 
u 
uv dx 
[uv] 
uvdx. 

Untitled picture.png Machine generated alternative text:
Using integration by parts, find the following indefinite integrals: 
5x 
ii) x + 
Note: the Maple notation for the exponential ex is exp (x) 
0 
o åt+C

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Find the definite integral 
25 
In x dc 
4 
Note: 
2/9 (-234 + - 
• the Maple notation for Inx is In (x) 
• you can use the preview button to preview your answer. 

Untitled picture.png Machine generated alternative text:
In some situations, you might have to apply integration by parts twice With this in mind, find 
sin(4 x) dc = -(xA2 * + + 
0åt+C 
Note: you may need to recall that the Maple notation for the exponential ex is exp (x) , or that the Maple notation for sinx is sin (x) 

Untitled picture.png Machine generated alternative text:
For the following, you might have to do something else first, before you get an integration by parts: 
dc = 2sin(sqrt(x)) - 
25 
1 
0åt+C 
Note: the Maple notation for is sqrt (x) , and and the Maple notation for e 
is exp (x) 

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