How do you evaluate the definite integral #int 15x^2(1+x^3)^4# from #[-1,1]#?

Question

Luke Alvarez · Accepted Answer

Use a #u#-substitution to get #int_-1^1 15x^2(1+x^3)^4dx=32#.

Begin by noting that since #15# is constant, we can pull it out of the integral, simplifying the problem down to: #15int_-1^1x^2(1+x^3)^4dx# Also note that the derivative of #x^3# is #x^2#, and we have an #x^2# in the integral. That makes this problem a textbook candidate for a #u#-substitution: Let #color(blue)u=color(blue)(1+x^3)# #u=1+(-1)^3=color(green)0-># for the boundaries of the integral #u=1+(1)^3=color(green)2-># for the boundaries of the integral #(du)/dx=3x^2# #color(red)(du)=color(red)(3x^2dx)# Using the substitution, #5int_color(green)0^color(green)2color(red)(3x^2)(color(blue)(1+x^3))^4color(red)dx=5int_color(green)0^color(green)2color(blue)u^4color(red)(du)# And this integral is a simple case of reverse power rule: #5int_0^2u^4du=[5(u^5/5)]_0^2=[u^5]_0^2# #color(white)(XX)=((2)^5-(0)^5)=32#

Paisley Johnson · Answer

undefined To evaluate the definite integral $\int_{-1}^{1} 15x^2(1+x^3)^4$, you can use the substitution method. Let $u = 1 + x^3$. Then, $du = 3x^2 dx$. Rearrange to solve for $dx$, getting $dx = \frac{1}{3x^2} du$. Now substitute $u$ and $dx$ into the integral.

$\int_{-1}^{1} 15x^2(1+x^3)^4 dx = \int_{u(-1)}^{u(1)} 15(1/3)(u)^4 du$.

Now integrate with respect to $u$, and then substitute $u = 1 + x^3$, and evaluate from -1 to 1.

After integration and substitution, the result will be a numerical value.