What is #int_(0)^(8) sqrt( 8x-x^(2))dx #?

Question

What is #int_(0)^(8)  sqrt( 8x-x^(2))dx #?

Christopher Allers · Accepted Answer

#I=8pi#

Here,

#I=int_0^8 sqrt(8x-x^2dx#

#=int_0^8sqrt(16-(x^2-8x+16))dx#

#=int_0^8sqrt(4^2-(x-4)^2)dx#

We know that ,

#color(red)(intsqrt(a^2-X^2)dX=X/2sqrt(a^2-X^2)+a^2/2sin^-1(X/a)+c#

Put, #a=4 and X=x-4#

#I=[(x-4)/4sqrt(4^2-(x-4)^2)+4^2/2sin^-1((x-4)/4)]_0^8#

#=>I=[(x-4)/4sqrt(8x-x^2)+8sin^-1((x-4)/4)]_0^8#

#=>I=[(8-4)/4sqrt(64-64)+8sin^-1((8-4)/4)]-[(0-4)/4sqrt(0- 0^2)+8sin^-1((0-4)/4)]#

#=[0+8sin^-1(1)]-[0+8sin^-1(-1)]#

#=8sin^-1(1)-8sin^-1(-1)#

#=8sin^-1(1)+8sin^-1(1)#

#=8(pi/2)+8(pi/2)#

#=4pi+4pi#

#=8pi#

Paisley Johnson · Answer

undefined To solve the integral $\int_{0}^{8} \sqrt{8x - x^2} \, dx$, we first complete the square inside the square root:

$$8x - x^2 = -(x^2 - 8x) = -(x^2 - 8x + 16 - 16) = -(x - 4)^2 + 16$$

So, $\sqrt{8x - x^2} = \sqrt{-(x - 4)^2 + 16}$.

Now, let $u = x - 4$, then $du = dx$. When $x = 0$, $u = -4$, and when $x = 8$, $u = 4$. So, the integral becomes:

$$\int_{-4}^{4} \sqrt{-u^2 + 16} \, du$$

This is a standard form of the semi-circle with radius 4, which can be evaluated using trigonometric substitution:

Let $u = 4 \sin(\theta)$, then $du = 4 \cos(\theta) d\theta$. When $u = -4$, $\theta = -\frac{\pi}{2}$, and when $u = 4$, $\theta = \frac{\pi}{2}$. Substituting into the integral:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4\sqrt{1 - \sin^2(\theta)} \cdot 4 \cos(\theta) \, d\theta$$

$$= 16\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta$$

Using the double-angle identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$, we get:

$$= 16\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta$$

$$= 16\left[\frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$

$$= 16\left[\frac{\pi}{4} + \frac{\sin(\pi)}{4} - \left(-\frac{\pi}{4} + \frac{\sin(-\pi)}{4}\right)\right]$$

$$= 16\left[\frac{\pi}{4} + \frac{\pi}{4}\right]$$

$$= 8\pi$$

So, $\int_{0}^{8} \sqrt{8x - x^2} \, dx = 8\pi$.