How do you evaluate the integral of #x^2# from 1 to 2?

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Answer 1 · 2016-02-18T21:25:33

#7/3#

Here, we have #f(x)=x^2#.

If #f# is continuous on #[a,b]# and #f(x)=F'(x)#, then

#int_a^bf(x)dx=F(b)-F(a)#

So, when we have

#int_1^2x^2dx#

we must find the antiderivative of the function #x^2# using the rule:

#intx^ndx=x^(n+1)/(n+1)+C#

Using this rule, we see that #F(x)# is #x^3/3#.

Thus,

#int_1^2x^2dx=F(2)-F(1)#

Note that #F(2)-F(1)=[x^3/3]_1^2#, this is just another way of writing it.

#F(2)-F(1)=2^3/3-1^3/3=8/3-1/3=7/3#

wolframalpha.com

This means that the area between the #x^2# and the #x#-axis on the interval #[1,2]# is #"7/3"#.