Question #9ae03

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Question #9ae03

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Answer 1 · 2017-02-05T06:52:09

$\frac{5}{8} {(\sqrt[5]{y^{4} - 8})}^{8} + C .$ $5/8(root5(y^4-8))^8+C.$ .

Explanation:

We will use the Substitution $: \sqrt[5]{y^{4} - 8} = x$ $: root5(y^4-8)=x$ .

$\Rightarrow (y^{4} - 8) = x^{5} \Rightarrow d (y^{4} - 8) = d (x^{5}), i . e ., 4 y^{3} d y = 5 x^{4} d x$ $rArr (y^4-8)=x^5 rArr d(y^4-8)=d(x^5), i.e., 4y^3dy=5x^4dx$ .

$Therefore, I = \int {(\sqrt[5]{y^{4} - 8})}^{3} (4 y^{3} d y)$ $"Therefore, "I=int(root5(y^4-8))^3(4y^3dy)$

$= \int (x^{3}) (5 x^{4}) d x = 5 \int x^{7} d x = 5 (\frac{x^{7 + 1}}{7 + 1}) = \frac{5}{8} x^{8}$ $=int(x^3)(5x^4)dx=5intx^7dx=5(x^(7+1)/(7+1))=5/8x^8$ .

$\Rightarrow I = \frac{5}{8} {(\sqrt[5]{y^{4} - 8})}^{8} + C .$ $rArr I=5/8(root5(y^4-8))^8+C.$ .

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Question #9ae03

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