What is #int_(0)^(2) x^(3)dx #?

1 Answer
Oct 28, 2015

4 sq units


Remember that an application of integration is to find the area under a curve. In this example you are finding the area bounded between 0 and 2.

First you apply the integration rules.

#int x^3dx=(x^4)/4#

Now apply the bounds of 0 and 2.


Substitute in the upper bound into the expression
Substitute in the lower bound into the expression
Subtract the lower bound result from the upper bound result

#[(x^4)/4]_0^2=(2^4)/4-(0^4)/4=16/4-0=16/4=4# square units