Answer to Question #178351 in Calculus for Phyroe

Question #178351

Integration Procedures (Integration by Parts)


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


1
Expert's answer
2021-04-13T23:57:29-0400

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" .


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