WEBVTT
1
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to evaluate the definite integral from 1 to 4 of
2
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x times F double prime of x dx. Given
3
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the information here, we would apply integration by parts
4
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and here we let u equal to X. And
5
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devi would be F double prime of X dx.
6
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And so the differential of you would be equal to
7
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dx And we would be the integral of deVI which
8
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is F prime of X. and so the integral
9
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from 1 to 4 of X times F double prime
10
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of x dx. This is just U v minus
11
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the integral of VD you. This would be extends
12
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F prime of x minus the integral of F prime
13
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of X. The X The first term evaluated from
14
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1 to 4 And we have this integral from 1-4
15
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, integrating further, we would have X times F
16
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prime of X Evaluated from 1 to 4-4 f
17
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of X, Evaluated from 1- four. Now
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we can combine these two and you would get x
19
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times f prime of x minus f of X,
20
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Evaluated from 1- four. When access for we
21
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have four times Ephraim A four-F of four and
22
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then when exist one, we would have one times
23
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F of one or f rank of one-F of
24
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one. This is equal to four times F.
25
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Prima for which is equal to three minus Fo four
26
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, which is seven and then minus. We have
27
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one times F prime of one, which is five
28
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-F of one, which is equal to two.
29
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Simplifying this, you would get 12-7-3 which
30
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is equal to two and so this is the value
31
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of the definite integral.