# Thread: Area enclosed by a curve?.

1. ## Area enclosed by a curve?.

Hey can anybody help me by giving a step to step guide on how to work out an area enclosed by a curve please?.

Find the area enclosed by the curve: Y=4x^2 (x^2-2x+1). The axis and the ordinates x=1 and x=2.

thanks

2. ## Re: Area enclosed by a curve?.

Originally Posted by MstrKurt
Hey can anybody help me by giving a step to step guide on how to work out an area enclosed by a curve please?.

Find the area enclosed by the curve: Y=4x^2 (x^2-2x+1). The axis and the ordinates x=1 and x=2.
I assume the area between the curve and the x-axis ... what do you know about definite integrals ?

$A = \int_1^2 4x^2(x^2-2x+1) \, dx$

3. ## Re: Area enclosed by a curve?.

I know that definite integrals have limits, and always end up with a numerical answer with no need for +c at the end.

4. ## Re: Area enclosed by a curve?.

I have worked out that the integrated equation would look like
4x^5-8x^4+4x^3
However, apparently the -8x^3 turns to -2x^4
Can anybody explain why?. Thanks.

5. ## Re: Area enclosed by a curve?.

$A = \int_1^2 4x^2(x^2-2x+1) \, dx$

$A = \int_1^2 4x^4 - 8x^3 + 4x^2 \, dx$

the antiderivative of $k x^n$ is $\frac{k x^{n+1}}{n+1}$

$\left[\frac{4x^5}{5} - 2x^4 + \frac{4x^3}{3}\right]_1^2$

$\left(\frac{128}{5} - 32 + \frac{32}{3}\right) - \left(\frac{4}{5} - 2 + \frac{4}{3}\right)$

I leave you to finish the arithmetic