Area enclosed by a curve?.

**MstrKurt** · May 14th 2012, 11:32 AM

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

**skeeter** · May 14th 2012, 11:55 AM

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 ?

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

**MstrKurt** · May 14th 2012, 12:01 PM

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

**MstrKurt** · May 14th 2012, 12:56 PM

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.

**skeeter** · May 14th 2012, 02:05 PM

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

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

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

Thread: Area enclosed by a curve?.

LinkBack

Thread Tools

Display

Area enclosed by a curve?.

Re: Area enclosed by a curve?.

Re: Area enclosed by a curve?.

Re: Area enclosed by a curve?.

Re: Area enclosed by a curve?.

Similar Math Help Forum Discussions

Determine the area enclosed by the curve y = cos ^2 fi

Area enclosed by a plane curve... and converse

area enclosed by curve

Help with area enclosed by curve

Area enclosed by a complex curve

Search Tags