Question

How to find area of these graphs via intergration?

Answer 1

I tried this buth check my maths anyway...

Explanation:

In exercise 4 we can think in terms of subtracting areas; we take a big area, normally a rectangle, and subtract smaller "curved" areas (the area of P for example is the composition of 2 smaller areas formed by a big rectangle minus a "curved" trapezium)

Have a look:

Answer 2

4,P)`A=int_1^e2lny*dy=[2xlnx-x]_1^e=2 (unite)^2`
4,Q)`A=int_0^1e^x-e^-x*dx= 1.0862 (unite)^2`
5)`A=int_0^(7pi/6)sin(x)-(-1/2)*dx=3.699 (unite)^2`

Explanation:

for question 4
a) for area P

`y=1and y=e`

`y=e^x`
solve for x
`x=lny`

`y=e^-x`
solve for x
`x=-lny`

`A=int_a^b(x_2-x_1)*dy=int_1^elny-(-lny)*dy`

`int_1^elny+lny*dy=int_1^e2lny*dy=[2xlnx-x]_1^e=2 (unite)^2`

b) for area Q

`x=0 and x=1`

`A=int_a^b(y_2-y_1)*dx=int_0^1e^x-e^-x*dx=[e^x+e^-x]_0^1=1.0862 (unite)^2`

For question 5

The area of shaded region

`x=0 and x=7pi/6`

`A=int_a^b(y_2-y_1)*dx=int_0^(7pi/6)sin(x)-(-1/2)*dx`

`[x/2-cos(x)]_0^(7pi/6)=(7*pi+2*3^(3/2)+12)/12=3.699 (unite)^2`