How do you evaluate the integral $\int_{0}^{1} x^{2} d x$ $int_0^1x^2dx$ ?

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Answer 1 · 2014-09-25T17:17:38

$\int_{0}^{1} x^{2} d x = \frac{1}{3}$ $int_0^1 x^2 dx = 1/3$

Remember one of the most important theorems in Calculus:

Fundamental Theorem of Calculus (Part 2)

$\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)$ $int_a^b f(x)dx=[F(x)]_a^b=F(b)-F(a)$ ,

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

Let us the theorem above to evaluate the definite integral.

$\int_{0}^{1} x^{2} d x$ $int_0^1 x^2 dx$

by finding an antiderivative of $x^{2}$ $x^2$ using Power Rule,

$= {[\frac{x^{3}}{3}]}_{0}^{1}$ $=[x^3/3]_0^1$

by plugging in the upper and the lower limits,

$= \frac{{(1)}^{3}}{3} - \frac{{(0)}^{3}}{3}$ $=(1)^3/3-(0)^3/3$

by simplifying,

$= \frac{1}{3}$ $=1/3$

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