# Thread: integration (area under a curve)

1. ## 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.

Thank you in advance! 2. 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

3. 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.

Thank you in advance! 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. $\displaystyle \frac{d}{dx} (2x-3)^3 = 6(2x-3)^2$

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

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

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 $\displaystyle \int_{0}^{\frac{3}{2}} (2x-3)^3dx$

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.

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?

5. 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 $\displaystyle 6(2x-3)^2$.
Then equal it to 6.

6. sorry, my mistake.
Thank you.

7. 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?

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

solve the last quadratic equation to find your 2 and the another one.

9. Just to point out that from Ted's quadratic you may find it easier to solve if you divide both sides by 4

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