help with integration question?

**decoy808** · May 19th 2010, 01:12 PM

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

**skeeter** · May 19th 2010, 02:42 PM

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.

**DrDank** · May 19th 2010, 02:49 PM

$<br /> <br />$

A and B occur when y=0

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

$<br /> 0=(x-1)$ 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

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