# integration (area under a curve)

• Feb 18th 2010, 07:56 AM
Oasis1993
integration (area under a curve)
Hello guys! Im having trouble with this question i would appreciate the help.

curve y=( 2x-3)^3

a) find the x coordinates of the two points on the curve which have gradient 6

b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.

Attachment 15478
• Feb 18th 2010, 08:12 AM
Ted
1- For (a) solve $\displaystyle y'=6$.
2- For (b) start by finding the L/U limits of the integration.
3- You should post this in the calculus sub-forum.
• Feb 20th 2010, 02:12 AM
HallsofIvy
Quote:

Originally Posted by Oasis1993
Hello guys! Im having trouble with this question i would appreciate the help.

curve y=( 2x-3)^3

a) find the x coordinates of the two points on the curve which have gradient 6

b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.

Attachment 15478

It would help if you would tell us what trouble you are having! As Ted told you, to do the first, you simply find the general formula for the derivative (gradient), set it equal to 6, and solve for x. Is the difficulty finding the deerivative of $\displaystyle (2x- 3)^3$? You will need the "power rule" and the "chain rule". Do you know those?

For (b), You appear to have left out important information. You cannot find the area bounded by a curve without know what curve! Are you not given a formula for the graph?
• Feb 20th 2010, 02:26 AM
Prove It
Quote:

Originally Posted by HallsofIvy
It would help if you would tell us what trouble you are having! As Ted told you, to do the first, you simply find the general formula for the derivative (gradient), set it equal to 6, and solve for x. Is the difficulty finding the deerivative of $\displaystyle (2x- 3)^3$? You will need the "power rule" and the "chain rule". Do you know those?

For (b), You appear to have left out important information. You cannot find the area bounded by a curve without know what curve! Are you not given a formula for the graph?

If you don't know the chain rule yet, you'll have to expand the expression first and then just use power rule.
• Feb 20th 2010, 07:54 AM
Hello Oasis1993
Quote:

Originally Posted by Oasis1993
Hello guys! Im having trouble with this question i would appreciate the help.

curve y=( 2x-3)^3

a) find the x coordinates of the two points on the curve which have gradient 6

b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.

Attachment 15478

(a)
$\displaystyle y = (2x-3)^3$
$\displaystyle \Rightarrow\frac{dy}{dx}= 6(2x-3)^2$
$\displaystyle =6$
when
$\displaystyle (2x-3)^2=1$

$\displaystyle \Rightarrow 2x-3 = \pm1$

$\displaystyle \Rightarrow x = 1, 2$
So the points at which the gradient is $\displaystyle 6$ are $\displaystyle (1,-1)$ and $\displaystyle (2,1)$.

(b) The curve $\displaystyle y=( 2x-3)^3$ cuts the $\displaystyle y$-axis when $\displaystyle x = 0$ (and $\displaystyle y = -27$) and touches the $\displaystyle x$-axis when $\displaystyle (2x-3)^3 = 0$; i.e. when $\displaystyle x = 1.5$. So the area between the curve and the axes is:
$\displaystyle \int_0^{1.5}(2x-3)^3dx$
$\displaystyle =\int_0^{1.5}(8x^3-36x^2+54x-27)\;dx$

$\displaystyle =\Big[2x^4-12x^3+27x^2-27x\Big]_0^{1.5}$

$\displaystyle = -\frac{81}{8}$, if my arithmetic is correct.
So the required area is $\displaystyle \frac{81}{8}$ sq. units.