Question

How to answer these using intergration ?

Answer 1

The area is `=(32/3)u^2` and the volume is `=(512/15pi)u^3`

Explanation:

Start by finding the intercept with the x-axis

`y=4x-x^2=x(4-x)=0`

Therefore,

`x=0` and `x=4`

The area is

`dA=ydx`

`A=int_0^4(4x-x^2)dx`

`=[2x^2-1/3x^3]_0^4`

`=32-64/3-0`

`=32/3u^2`

The volume is

`dV=piy^2dx`

`V=piint_0^4(4x-x^2)^2dx`

`=piint_0^4(16x^2-8x^3+x^4)dx`

`=pi[16/3x^3-2x^4+1/5x^5]_0^4`

`=pi(1024/3-512+1024/5-0)`

`=pi(5120/15-7680/15+3072/15)`

`=pi(512/15)`

Answer 2

a. `32/3`

b. `(512pi)/15`

Explanation:

First, we need to find the points at which the graph crosses the `x`-axis.

`4x-x^2=x(4-x)=0`

Either `x=0` or `4-x=0`

`x=0 or 4`

Now we know our upper and lower bounds.

a. `"Area under a graph"=int_b^af(x)dx`

`int_0^4 4x-x^2dx=[2x^2-x^3/3]_0^4=(2(4)^2-4^3/3)-(2(0)^2-0^3/3)=32/3`

b. `"Volume of rotation"=piint_b^a(f(x))^2dx`

`f(x)^2=(4x-x^2)^2=16x^2-8x^3+x^4`

`piint_0^4 16x^2-8x^3+x^4dx=pi[(16x^3)/3-2x^4+x^5/5]_0^4=pi[((16(4)^3)/3-2(4)^4+4^5/5)-((16(0)^3)/3-2(0)^4+0^5/5)]=pi[512/15]=(512pi)/15`