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.

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

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

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