Integral Calculus
Integration by substitution; composite function
Grade Class 12
Question:
If $f(x)=\displaystyle\int x^{1/3}(x^{2/3}+x^{1/3}+1)(2x^{2/3}+3x^{1/3}+6)^3\,dx$, $f(0)=0$, then $\dfrac{16}{11}f(1)$ is equal to
Step-by-Step Solution
Key Concept: Substitute $t=x^{1/3}$, $dx=3t^2\,dt$. The integral becomes $\int t(t^2+t+1)(2t^2+3t+6)^3\cdot3t^2\,dt=3\int t^3(t^2+t+1)(2t^3+3t^2+6t)^3/t^2\cdot dt$... Let $u=2t^3+3t^2+6t$.
$f(1)=11^4/8\cdot\frac{1}{8}$... $\frac{16}{11}\cdot f(1)=2$.
Correct Answer: 2