Hello Oasis1993 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!

(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.

Grandad