Question

How do you solve the following questions?

Answer 1

i) You should notice that

`A_"shaded"= A_"rectangle" - A_"under root graph"`

`A_"shaded" = 2 - int_0^1 sqrt(3x+ 1) dx`

The integral can be evaluated thru a simple u-substitution. Let `u = 3x + 1`. Then `du = 3dx` and `dx = (du)/3`.

`A_"shaded" = 2- 1/3int_1^4 sqrt(u) du`

`A_"shaded" = 2 - 1/3[2/3u^(3/2)]_1^4`

`A_"shaded" = 2 - 1/3[2/3(4)^(3/2) - 2/3(1)^(3/2)]`

`A_"shaded" = 2 - 1/3(16/3 - 2/3)`

`A_"shaded" = 2 - 14/9`

`A_"shaded" = 4/9`

ii) What we essentially have to do is find the volume if the area under the line `y= 2` on `[0, 1]` is rotated around the x-axis and then subtract the volume if the area under `y = sqrt(3x + 1)` on `[0, 1]` is rotated around the x-axis.

`V = piint_0^1 2^2dx - pi int_0^1 (sqrt(3x + 1))^2 dx`

`V = pi[4x]_0^1 - pi[3/2x^2 + x]_0^1 dx`

`V = 4pi - pi(3/2 + 1)`

`V = 4pi - 3/2pi - pi`

`V = 3/2 pi`

`V = 0.477`

iii) I will let another contributor do this one.

Hopefully this helps!