(int_(0)^(a)4x^(4)sqrt(a^(2)-x^(2))dx)/(int_(0)^(a)(ax)^(2)sqrt(a^(2)-x^(2))dx)=

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Jul 02, 2024

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Evaluating the Definite Integral

In this article, we will evaluate the following definite integral:

$\frac{\int_{0}^{a} 4x^4 \sqrt{a^2 - x^2} dx}{\int_{0}^{a} ax^2 \sqrt{a^2 - x^2} dx}$

To evaluate this integral, we will first simplify the numerator and denominator separately.

Numerator

Let's evaluate the numerator:

$\int_{0}^{a} 4x^4 \sqrt{a^2 - x^2} dx$

We can start by substituting $x = a \sin \theta$, which implies that $dx = a \cos \theta d\theta$. Then, we have:

$\int_{0}^{a} 4x^4 \sqrt{a^2 - x^2} dx = \int_{0}^{\frac{\pi}{2}} 4a^4 \sin^4 \theta \sqrt{a^2 - a^2 \sin^2 \theta} a \cos \theta d\theta$

Simplifying, we get:

$\int_{0}^{\frac{\pi}{2}} 4a^5 \sin^4 \theta \cos^2 \theta d\theta$

Now, we can use the following trigonometric identities:

• $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$
• $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$

Substituting these identities, we get:

$\int_{0}^{\frac{\pi}{2}} 4a^5 \left( \frac{1 - \cos 2\theta}{2} \right)^2 \left( \frac{1 + \cos 2\theta}{2} \right) d\theta$

Simplifying and integrating, we get:

$\int_{0}^{\frac{\pi}{2}} 4a^5 \left( \frac{3 - 2\cos 2\theta - \cos^2 2\theta}{8} \right) d\theta$

Evaluating the integral, we get:

$\frac{a^5 \pi}{8}$

Denominator

Now, let's evaluate the denominator:

$\int_{0}^{a} ax^2 \sqrt{a^2 - x^2} dx$

Using the same substitution $x = a \sin \theta$, we get:

$\int_{0}^{a} ax^2 \sqrt{a^2 - x^2} dx = \int_{0}^{\frac{\pi}{2}} a^3 \sin^2 \theta \sqrt{a^2 - a^2 \sin^2 \theta} a \cos \theta d\theta$

Simplifying, we get:

$\int_{0}^{\frac{\pi}{2}} a^4 \sin^2 \theta \cos^2 \theta d\theta$

Using the same trigonometric identities, we get:

$\int_{0}^{\frac{\pi}{2}} a^4 \left( \frac{1 - \cos 2\theta}{2} \right) \left( \frac{1 + \cos 2\theta}{2} \right) d\theta$

Simplifying and integrating, we get:

$\int_{0}^{\frac{\pi}{2}} a^4 \left( \frac{1 - \cos^2 2\theta}{4} \right) d\theta$

Evaluating the integral, we get:

$\frac{a^4 \pi}{8}$

Final Answer

Now, we can evaluate the original expression:

$\frac{\int_{0}^{a} 4x^4 \sqrt{a^2 - x^2} dx}{\int_{0}^{a} ax^2 \sqrt{a^2 - x^2} dx} = \frac{\frac{a^5 \pi}{8}}{\frac{a^4 \pi}{8}} = \boxed{a}$

Therefore, the final answer is $a$.