# integration (area under a curve)

• Feb 18th 2010, 08:17 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.

http://www.mathhelpforum.com/math-he...-maths-pic.jpg
• Feb 18th 2010, 08:34 AM
Calculus26
a. f ' = 6(x-3)^2

now simply se this equal to 6

b. the integration limits are 0 and the x-intercept (set f = 0 simple enough)

the area is a naegative of the integral of f(x) on this interval
• Feb 18th 2010, 08:39 AM
Shananay
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.

http://www.mathhelpforum.com/math-he...-maths-pic.jpg

I see Calculus26 already answered you, but I wrote this all out so I'll just post it anyway.

a) Take the derivative a set it equal to 6. $\frac{d}{dx} (2x-3)^3 = 6(2x-3)^2$

solve for x when $6(2x-3)^2 = 6$

b) You are integrating $\int_{0}^{x} (2x-3)^3dx$

To find the upper limit of integration you need to set your function equal to zero, because that is the x value of your x intercept. Your function equals 0 when (2x-3) = 0, or when x = 3/2

So your integral becomes $\int_{0}^{\frac{3}{2}} (2x-3)^3dx$

You'll get a negative number for the area since the curve is below the x-axis on the interval which you're integrating, so your area will be the absolute value of the answer you get by evaluating that integral.
• Feb 18th 2010, 09:21 AM
Oasis1993
But could you please show the steps because i cant get the correct answer...
when i take the differentiaton of y=(2x-3)^3 i get 2(2x-3)^2 do i then have to equal it to 6?
• Feb 18th 2010, 09:27 AM
Ted
Quote:

Originally Posted by Oasis1993
But could you please show the steps because i cant get the correct answer...
when i take the differentiaton of y=(2x-3)^3 i get 2(2x-3)^2 do i then have to equal it to 6?

The derivative is $6(2x-3)^2$.
Then equal it to 6.
• Feb 18th 2010, 09:29 AM
Oasis1993
sorry, my mistake.
Thank you.
• Feb 19th 2010, 12:09 PM
Oasis1993
Ted,
can i ask one more question?
when i do 6(2x-3)^2 = 6
i get 2. which is correct but it asks for TWO x-coordinates.
how can i get the second one?
• Feb 19th 2010, 01:54 PM
Ted
Quote:

Originally Posted by Oasis1993
Ted,
can i ask one more question?
when i do 6(2x-3)^2 = 6
i get 2. which is correct but it asks for TWO x-coordinates.
how can i get the second one?

$6(2x-3)^2=6$.
devide both sides by 6:
$(2x-3)^2=1$.
expand:
$4x^2-12x+9=1$
$4x^2-12x+8=0$

solve the last quadratic equation to find your 2 and the another one.
• Feb 19th 2010, 01:58 PM
e^(i*pi)
Just to point out that from Ted's quadratic you may find it easier to solve if you divide both sides by 4