Question

How to solve with integration?

how to do question 10 a) b) and question 3 and 4?

Answer 1

`Q=(15/2,0)`
`P=(3,9)`
`"Area"=117/4`

Explanation:

Q is the x-intercept of the line `2x+y=15`

To find this point, let `y=0`
`2x=15`
`x=15/2`

So `Q=(15/2,0)`

P is a point of interception between the curve and the line.

`y=x^2" "(1)`
`2x+y=15" "(2)`

Sub `(1)` into `(2)`

`2x+x^2=15`
`x^2+2x-15=0`
`(x+5)(x-3)=0`
`x=-5` or `x=3`

From the graph, the x co-ordinate of P is positive, so we can reject `x=-5`

`x=3`
`y=x^2`
`=3^2`
`=9`

`:. P=(3,9)`

graph{(2x+y-15)(x^2-y)=0 [-17.06, 18.99, -1.69, 16.33]}

Now for the area

To find the total area of this region, we can find two areas and add them together.
These will be the area under `y=x^2` from 0 to 3, and the area under the line from 3 to 15/2.

`"Area under curve"=int_0^3 x^2dx`
`=[1/3x^3]_0^3`
`=1/3xx3^3-0`
`=9`

We can work out the area of the line through integration, but its easier to treat it like a triangle.

`"Area under line"=1/2xx9xx(15/2-3)`
`=1/2xx9xx9/2`
`=81/4`

`:."total area of shaded region"=81/4+9`
`=117/4`

Answer 2

For 3 & 4

[Tom's done 10]

Explanation:

3

`int_0^5 f(x) \ dx = (int_0^1 + int_1^5) f(x) \ dx`

`:. int_1^5 f(x) \ dx = (int_0^5 - int_0^1) f(x) \ dx`

`= 1- (-2) = 3`

4

`int_(-2)^3 f(x) dx = (int_(-2)^1 + int_1^3) f(x) dx`

`:. int_(3)^(-2) f(x) dx = -int_(-2)^3 f(x) dx`

`= - (int_(-2)^1 + int_1^3) f(x) dx`

`= - (2 - 6) = 4`

Answer 3

See below:

Warning: Long answer!

Explanation:

For (3):

Using the property:

`int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx`

Hence:

`int_0^5 f(x) dx=int_0^1 f(x) dx + int_1^5 f(x) dx`

`1=-2 + x`

`x=3=int_1^5 f(x) dx`

For (4):

(same thing)

`int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx`

`int_-2^3 f(x) dx=int_-2^1 f(x) dx + int_1^3 f(x) dx`

`x=2+(-6)`

`x=-4=int_-2^3 f(x) dx`

However, we must swap the limits on the integral, so:
`int_3^-2 f(x) dx=-int_-2^3 f(x) dx`

So:`int_3^-2 f(x) dx=-(-4)=4`

For 10 (a):

We have two functions intersecting at `P`, so at `P`:

`x^2=-2x+15`

(I turned the line function into slope-intercept form)

`x^2 + 2x-15=0`

`(x+5)(x-3)=0`

So `x=3` as we to the right of the `y` axis, so `x>0`.

(inputting `x=3` into any of the functions)

`y=-2x+15`

`y=-2(3)+15`

`y=15-6=9`

So the coordinate of `P` is `(3,9)`

For `Q`, the line `y=-2x+15` cuts the `y`-axis, so `y=0`

`0=-2x+15`

`2x=15`

`x=(15/2)=7.5`

So `Q` is located at `(7.5, 0)`

For 10 (b).

I will construct two integrals to find the area. I will solve the integrals separately.

The area is:
`int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx`

`A=int_O^Q f(x) dx=int_O^P (x^2) dx + int_P^Q (-2x+15) dx`

(Solve first integral)

`int_O^P (x^2) dx=int_0^3 (x^2) dx=[x^3/3]`

(substitute the limits into the integrated expression, remember:
Upper-lower limit to find the value of integral)

`[3^3/3]-0=9=int_O^P (x^2) dx`

(solve second integral)

`int_P^Q (-2x+15) dx=int_3^7.5 (-2x+15) dx=[(-2x^2)/2+15x]=[-x^2+15x]`

(substitute limits: Upper-lower)

`[-(15/2)^2+15(15/2)]-[-3^2+15(3)]`

#[(-225/4)+(225/2)]+[9-45]=[(-225/4)+(450/4)]+[-36]= [(225/4)]+[(-144/4)]=(81/4)#

`int_P^Q (-2x+15)dx=(81/4)`

`int_O^Q f(x) dx=int_O^P (x^2) dx + int_P^Q (-2x+15) dx`

`A=int_O^Q f(x) dx=9 + (81/4)`

`A=int_O^Q f(x) dx=9 + (81/4)`

`A=(36/4)+(81/4)`

`A=(117/4)`