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.

1 thought on “Evaluate a Definite Integral without FTC

  1. Pingback: From Dancing Planet to Kepler’s Laws | Vroom

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