if-f-x-3-1-x-5-4x-2-find-0-1-f-x-dx-

Question Number 163891 by HongKing last updated on 11/Jan/22

if f (x^{3} + 1) = x^{5} + 4 x + 2

find \underset{0}{\int^{1}} f (x) dx = ?

Answered by Ar Brandon last updated on 11/Jan/22

x^{3} + 1 = y \Rightarrow x = \sqrt[3]{y - 1}

\Rightarrow f (y) = {(y - 1)}^{\frac{5}{3}} + 4 {(y - 1)}^{\frac{1}{3}} + 2

\Rightarrow f (x) = {(x - 1)}^{\frac{5}{3}} + 4 {(x - 1)}^{\frac{1}{3}} + 2

\int_{0}^{1} f (x) d x = {[\frac{3}{8} {(x - 1)}^{\frac{8}{3}} + 3 {(x - 1)}^{\frac{4}{3}} + 2 x]}_{0}^{1}

= 2 - (\frac{3}{8} + 3) = - \frac{11}{8}

Commented by HongKing last updated on 11/Jan/22

cool my dear Sir thank you so much

Answered by mr W last updated on 11/Jan/22

\int_{0}^{1} f (x) dx

= \int_{- 1}^{0} f (t^{3} + 1) d (t^{3} + 1)

= \int_{- 1}^{0} {3 t}^{2} (t^{5} + 4 t + 2) dt

= 3 {[\frac{t^{8}}{8} + t^{4} + \frac{{2 t}^{3}}{3}]}_{- 1}^{0}

= 3 (- \frac{1}{8} - 1 + \frac{2}{3})

= - \frac{11}{8}

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