# Evaluate a Definite Integral without FTC

Problem: Find the area enclosed by ellipse

$\frac{x^2}{a^2} + \frac{y^2}{b^2}= 1$

Solution: For $0 \le x \le a$,

$y = b\cdot\sqrt{1-\frac{x^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}$.

Clearly,

$A_{\;\frac{1}{4}ellipse} =\int\limits_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}\;dx =\frac{b}{a}\cdot\boxed{\int\limits_{b}^{a}\sqrt{a^2-x^2}\;dx}$.

Since $\int\limits_{0}^{a}\sqrt{a^2-x^2}\;dx$ is a quarter of the area enclosed by a circle with radius $a$,

$\int\limits_{0}^{a}\sqrt{a^2-x^2}\;dx = \frac{\pi a^2}{4}$.

It follows that

$A_{\; \frac{1}{4} ellipse} =\frac{b}{a}\cdot \frac{\pi a^2}{4}=\frac{\pi a b }{4} \implies A_{\;ellipse} = \pi a b$.