# Evaluate the integral. {eq}\displaystyle \int_0^2 \sqrt {4 - x^2}\ dx {/eq}

## Question:

Evaluate the integral.

{eq}\displaystyle \int_0^2 \sqrt {4 - x^2}\ dx {/eq}

## Definite Integral:

The definite integral {eq}\displaystyle \int_{c}^{d} g(x)\; dx {/eq} is a number; it does not depend on {eq}x {/eq}. In fact, we could use any letter in place of {eq}x {/eq} without changing the value of the integral {eq}\displaystyle \int_{c}^{d} g(x) \; dx = \int_{c}^{d} g(t) \; dt = \int_{c}^{d} g(r) \; dr. {/eq}

We are given

{eq}\displaystyle \int_0^2 \sqrt {4 - x^2}\ dx {/eq}

Perform the definite integral:

{eq}\begin{align} \displaystyle \implies \int_0^2 \sqrt {4 - x^2}\ dx &= \int_{0}^{2} \sqrt{ (2)^2 - x^2 } \; dx\\ &= \left[ \dfrac{x \; \sqrt{4-x^2}}{2}+ \dfrac{4}{2} \sin^{-1} \left(\dfrac{x}{2} \right) \right]_{0}^{2},\; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[ \int \sqrt{a^2 -x^2} = \dfrac{x\sqrt{a^2 -x^2}}{2} +\dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a} \right) + c \right] \\ &= \left[ \dfrac{x \; \sqrt{4 - x^2}}{2} + 2 \sin^{-1} \left( \dfrac{x}{2} \right) \right]_{0}^{2}\\ &= 0 + 2 \sin^{-1} (1) -0 -2 \sin^{-1} (0) \\ &= 2 \cdot \dfrac{\pi}{2}\\ &= \pi\\ \end{align} {/eq}