# ∫√2^1 (u^7/2 - 1/u^5)du

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### 2 Answers

You need to use the property of linearity of integral, hence, you need to split the given integral in two such that:

`int_1^sqrt2 (u^7/2 - 1/u^5)du = int_1^sqrt2 (u^7/2)du - int_1^sqrt2 (1/u^5)du `

`int_1^sqrt2 (u^7/2 - 1/u^5)du = (1/2)int_1^sqrt2 u^7 du -int_1^sqrt2 u^(-5) du`

`int_1^sqrt2 (u^7/2 - 1/u^5)du = (1/2)(u^8/8)|_1^sqrt2 - (u^(-5+1))/(-5+1)|_1^sqrt2 `

`int_1^sqrt2 (u^7/2 - 1/u^5)du = (1/2)(u^8/8)|_1^sqrt2 + 1/(4u^4)|_1^sqrt2 `

You need to use the fundamental theorem of calculus such that:

`int_1^sqrt2 (u^7/2 - 1/u^5)du = (1/2)((sqrt2)^8/8 - 1^8/8)+ (1/4)(1/(sqrt2)^4 - 1/1^4)`

`int_1^sqrt2 (u^7/2 - 1/u^5)du = (1/2)(16/8 - 1/8) + (1/4)(1/4 - 1)`

`int_1^sqrt2 (u^7/2 - 1/u^5)du = 15/16- 3/16`

`int_1^sqrt2 (u^7/2 - 1/u^5)du = 12/16 => int_1^sqrt2 (u^7/2 - 1/u^5)du = 3/4`

**Hence, evaluating the given definite integral yields `int_1^sqrt2 (u^7/2 - 1/u^5)du = 3/4.` **

Definate integral of in the question above, sorry. Thank you!