How do you integrate $y = {(- 7 x^{4})}^{\frac{8}{5}}$ $y=(-7x^4)^(8/5)$ ?

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How do you integrate $y = {(- 7 x^{4})}^{\frac{8}{5}}$ $y=(-7x^4)^(8/5)$ ?

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Answer 1 · 2016-07-04T12:37:17

$- \frac{35}{37} x^{\frac{37}{5}} + c$ $-35/37x^(37/5)+c$

Explanation:

Require to use $power rule for integration$ $color(blue)"power rule for integration"$

$\int (a x^{n}) = \frac{a}{n + 1} x^{n + 1}$ $int(ax^n)=a/(n+1)x^(n+1)$

$\Rightarrow \int {(- 7 x^{4})}^{\frac{8}{5}} d x = \int - 7 x^{\frac{32}{5}} d x$ $rArrint(-7x^4)^(8/5)dx=int-7x^(32/5)dx$

applying the power rule.

$\Rightarrow \int - 7 x^{\frac{32}{5}} d x = \frac{- 7}{\frac{37}{5}} x^{\frac{37}{5}} + c = - \frac{35}{37} x^{\frac{37}{5}} + c$ $rArrint-7x^(32/5)dx=(-7)/(37/5)x^(37/5)+c=-35/37x^(37/5)+c$

where c is the constant of integration.

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How do you integrate $y = {(- 7 x^{4})}^{\frac{8}{5}}$ $y=(-7x^4)^(8/5)$ ?

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How do you integrate y=(-7x^4)^(8/5)y=(−7x4)85y=(-7x^4)^(8/5)?

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Explanation:

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How do you integrate $y = {(- 7 x^{4})}^{\frac{8}{5}}$ $y=(-7x^4)^(8/5)$ ?