Question

I must get my math grade up and I need help. Please help me, I am unfamiliar with the material and need explaining/Thanks so much!

Answer 1

EXPLANATION

$\mathrm{The\: area\: between\: curves\: is\: the\: area\: between\: a\: curve}\: f\mleft(x\mright)\: \mathrm{and\: a\: curve}\: g\mleft(x\mright)\: \mathrm{on\: an\: interval}\: \mleft[a,\: b\mright]\: \mathrm{given\: by}$

We can see that the needed area is defined between x=0 and x=6, and is below the curve y= sin^-1 (x/6)

First, computing the area below the curve y= sin^-1 (x/6):

Applying integration by parts:

$\int \: uv^{\prime}=uv-\int \: u^{\prime}v$$u=\arcsin \mleft(\frac{x}{6}\mright)$$v^{\prime}=1$$\frac{d}{dx}\mleft(\arcsin \mleft(\frac{x}{6}\mright)\mright)$

Apply the chain rule:

$\frac{df(u)}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$$f=\arcsin \mleft(u\mright),\: \: u=\frac{x}{6}$$=\frac{d}{du}\mleft(\arcsin \mleft(u\mright)\mright)\frac{d}{dx}\mleft(\frac{x}{6}\mright)$$\frac{d}{du}\mleft(\arcsin \mleft(u\mright)\mright)$

Apply the common derivative:

$\frac{d}{du}(\arcsin (u))=\frac{1}{\sqrt[]{1-u^2}}$$=\frac{1}{\sqrt{1-u^2}}$$=\frac{1}{\sqrt{1-u^2}}\frac{d}{dx}\mleft(\frac{x}{6}\mright)$$\mathrm{Substitute\: back}\: u=\frac{x}{6}$$=\frac{1}{\sqrt{1-\left(\frac{x}{6}\right)^2}}\frac{d}{dx}\mleft(\frac{x}{6}\mright)$

Taking the constant out and applying the common derivative:

$=\frac{1}{\sqrt{1-\left(\frac{x}{6}\right)^2}}\frac{d}{dx}\mleft(\frac{x}{6}\mright)$