Answer to Question #319727 in Real Analysis for Jyo

Question

Use the substitution theorem evaluate,

Integral 0 to 2 t^2[1+t^3]^-1/2dt

Answer 1

$I=\int_0^2 \frac{t^2}{\sqrt{1+t^3}}dt$

Substitution:

$u=t^3+1$ ; $du=3t^2dt$;

$u_1=0^3+1=1$ ; $u_2=2^3+1=9$.

$I=\int_0^2 \frac{t^2}{\sqrt{1+t^3}}dt=\frac13\int_0^2 \frac{3t^2dt}{\sqrt{1+t^3}}=\frac13\int_1^9 \frac{du}{\sqrt{u}}=$$\frac23\sqrt u|_1^9=\frac23(3-1)=\frac43$