How do you evaluate the integral #int_0^1x^2dx# ?

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Answer 1 · 2014-09-25T17:17:38

#int_0^1 x^2 dx = 1/3#

Remember one of the most important theorems in Calculus:

Fundamental Theorem of Calculus (Part 2)

#int_a^b f(x)dx=[F(x)]_a^b=F(b)-F(a)#,

where #F(x)# is an antiderivative of #f(x)#.

Let us the theorem above to evaluate the definite integral.

#int_0^1 x^2 dx#

by finding an antiderivative of #x^2# using Power Rule,

#=[x^3/3]_0^1#

by plugging in the upper and the lower limits,

#=(1)^3/3-(0)^3/3#

by simplifying,

#=1/3#