# help with integration question?

#### decoy808

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

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

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

#### skeeter

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

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

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

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

#### DrDank

$$\displaystyle$$

A and B occur when y=0

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

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

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

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

Limits of integration are 1 to 4