Question

Integration Procedures (Integration by Parts)

∫ x³dx/cuberoot of (8-x²) from 0 to √7

Answer 1

Integration Procedures (Integration by Parts)

∫ x³dx/cuberoot of (8-x²) from 0 to √7

Solution:

$I=\int_0^{\sqrt7}\frac{x^3}{\sqrt[3]{8-x^2}}dx=\frac12\int_0^{\sqrt7}\frac{x^2}{\sqrt[3]{8-x^2}}dx^2$ $=\frac12\int_0^{7}\frac{t}{\sqrt[3]{8-t}}dt$

Integration by parts:

$u=t$ ; $du=dt$ ;

$dv=\frac{dt}{\sqrt[3]{8-t}}$ ; $v=-\frac32(8-t)^{\frac23}$ .

$I=-\frac34\cdot t(8-t)^{\frac23}|_0^7+\frac12\int_0^7\frac32(8-t)^{\frac23}dt=-\frac34\cdot7-$ $\frac34\cdot\frac35(8-t)^{\frac53}|_0^7=-\frac{21}{4}-\frac{9}{20}+\frac{9\cdot2^5}{20}=\frac{87}{10}=8.7$

Answer: $\int_0^{\sqrt7}\frac{x^3}{\sqrt[3]{8-x^2}}dx=8.7$ .