# Thread: help with integration question?

1. ## help with integration question?

can anyone help me with the process to the question. points i need to research rather than the answers. many thanks

The function $y=2(x-1)(x-4)^2$

(a) find the values of A and B
(b) what is the size of the shaded part

2. Originally Posted by decoy808
can anyone help me with the process to the question. points i need to research rather than the answers. many thanks

The function $y=2(x-1)(x-4)^2$

(a) find the values of A and B
(b) what is the size of the shaded part

should be clear that A = 1 and B = 4 (roots of the cubic, one w/ multiplicity two)

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

expand the cubic, then integrate and evaluate the definite integral using the fundamental theorem of calculus.

3. $

$

A and B occur when y=0

$y=2(x-1)(x-4)^2$
$0=2(x-1)(x-4)^2$

$
0=(x-1)$
and $0=(x-4)$

Therefore...
$A=1$ and $B=4$

The Area is..
$\int2(x-1)(x-4)^2dx$
$2\int(x-1)(x^2-8x+16)dx$
$2\int(x^3-7x^2+24x-16)dx$

Limits of integration are 1 to 4