7/25/2019 WolframAlpha--Definite Integral Definite Integral 2016 01-17-0838

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mpute the definite integral:

5

6

125x 53 x

ctor out constants:

6

125

0

5x 53 x

r the integrandx 53, substitute u x 5 and u x.is gives a new lower bound u 0 5 5 and upper bound u 5 5 0:

6

125

5

0

u3

u

ply the fundamental theorem of calculus.

e antiderivative of u3 isu

4

4:

3 u4

2505

0

aluate the antiderivative at the limits and subtract.

3 u4

2505

0

3 04

250

3

25054 15

2:

Answer:

15

2