# How do you solve the following questions?

##### 1 Answer

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

#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

#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!