# How to solve with integration?

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

May 13, 2018

$Q = \left(\frac{15}{2} , 0\right)$
$P = \left(3 , 9\right)$
$\text{Area} = \frac{117}{4}$

#### Explanation:

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

To find this point, let $y = 0$
$2 x = 15$
$x = \frac{15}{2}$

So $Q = \left(\frac{15}{2} , 0\right)$

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

$y = {x}^{2} \text{ } \left(1\right)$
$2 x + y = 15 \text{ } \left(2\right)$

Sub $\left(1\right)$ into $\left(2\right)$

$2 x + {x}^{2} = 15$
${x}^{2} + 2 x - 15 = 0$
$\left(x + 5\right) \left(x - 3\right) = 0$
$x = - 5$ or $x = 3$

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

$x = 3$
$y = {x}^{2}$
$= {3}^{2}$
$= 9$

$\therefore P = \left(3 , 9\right)$

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.

$\text{Area under curve} = {\int}_{0}^{3} {x}^{2} \mathrm{dx}$
$= {\left[\frac{1}{3} {x}^{3}\right]}_{0}^{3}$
$= \frac{1}{3} \times {3}^{3} - 0$
$= 9$

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

$\text{Area under line} = \frac{1}{2} \times 9 \times \left(\frac{15}{2} - 3\right)$
$= \frac{1}{2} \times 9 \times \frac{9}{2}$
$= \frac{81}{4}$

$\therefore \text{total area of shaded region} = \frac{81}{4} + 9$
$= \frac{117}{4}$

May 13, 2018

For 3 & 4

[Tom's done 10]

#### Explanation:

3

${\int}_{0}^{5} f \left(x\right) \setminus \mathrm{dx} = \left({\int}_{0}^{1} + {\int}_{1}^{5}\right) f \left(x\right) \setminus \mathrm{dx}$

$\therefore {\int}_{1}^{5} f \left(x\right) \setminus \mathrm{dx} = \left({\int}_{0}^{5} - {\int}_{0}^{1}\right) f \left(x\right) \setminus \mathrm{dx}$

$= 1 - \left(- 2\right) = 3$

4

${\int}_{- 2}^{3} f \left(x\right) \mathrm{dx} = \left({\int}_{- 2}^{1} + {\int}_{1}^{3}\right) f \left(x\right) \mathrm{dx}$

$\therefore {\int}_{3}^{- 2} f \left(x\right) \mathrm{dx} = - {\int}_{- 2}^{3} f \left(x\right) \mathrm{dx}$

$= - \left({\int}_{- 2}^{1} + {\int}_{1}^{3}\right) f \left(x\right) \mathrm{dx}$

$= - \left(2 - 6\right) = 4$

May 13, 2018

See below:

#### Explanation:

For (3):

Using the property:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\int}_{a}^{c} f \left(x\right) \mathrm{dx} + {\int}_{c}^{b} f \left(x\right) \mathrm{dx}$

Hence:

${\int}_{0}^{5} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} f \left(x\right) \mathrm{dx} + {\int}_{1}^{5} f \left(x\right) \mathrm{dx}$

$1 = - 2 + x$

$x = 3 = {\int}_{1}^{5} f \left(x\right) \mathrm{dx}$

For (4):

(same thing)

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\int}_{a}^{c} f \left(x\right) \mathrm{dx} + {\int}_{c}^{b} f \left(x\right) \mathrm{dx}$

${\int}_{-} {2}^{3} f \left(x\right) \mathrm{dx} = {\int}_{-} {2}^{1} f \left(x\right) \mathrm{dx} + {\int}_{1}^{3} f \left(x\right) \mathrm{dx}$

$x = 2 + \left(- 6\right)$

$x = - 4 = {\int}_{-} {2}^{3} f \left(x\right) \mathrm{dx}$

However, we must swap the limits on the integral, so:
${\int}_{3}^{-} 2 f \left(x\right) \mathrm{dx} = - {\int}_{-} {2}^{3} f \left(x\right) \mathrm{dx}$

So:${\int}_{3}^{-} 2 f \left(x\right) \mathrm{dx} = - \left(- 4\right) = 4$

For 10 (a):

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

${x}^{2} = - 2 x + 15$

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

${x}^{2} + 2 x - 15 = 0$

$\left(x + 5\right) \left(x - 3\right) = 0$

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

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

$y = - 2 x + 15$

$y = - 2 \left(3\right) + 15$

$y = 15 - 6 = 9$

So the coordinate of $P$ is $\left(3 , 9\right)$

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

$0 = - 2 x + 15$

$2 x = 15$

$x = \left(\frac{15}{2}\right) = 7.5$

So $Q$ is located at $\left(7.5 , 0\right)$

For 10 (b).

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

The area is:
${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\int}_{a}^{c} f \left(x\right) \mathrm{dx} + {\int}_{c}^{b} f \left(x\right) \mathrm{dx}$

$A = {\int}_{O}^{Q} f \left(x\right) \mathrm{dx} = {\int}_{O}^{P} \left({x}^{2}\right) \mathrm{dx} + {\int}_{P}^{Q} \left(- 2 x + 15\right) \mathrm{dx}$

(Solve first integral)

${\int}_{O}^{P} \left({x}^{2}\right) \mathrm{dx} = {\int}_{0}^{3} \left({x}^{2}\right) \mathrm{dx} = \left[{x}^{3} / 3\right]$

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

$\left[{3}^{3} / 3\right] - 0 = 9 = {\int}_{O}^{P} \left({x}^{2}\right) \mathrm{dx}$

(solve second integral)

${\int}_{P}^{Q} \left(- 2 x + 15\right) \mathrm{dx} = {\int}_{3}^{7.5} \left(- 2 x + 15\right) \mathrm{dx} = \left[\frac{- 2 {x}^{2}}{2} + 15 x\right] = \left[- {x}^{2} + 15 x\right]$

(substitute limits: Upper-lower)

$\left[- {\left(\frac{15}{2}\right)}^{2} + 15 \left(\frac{15}{2}\right)\right] - \left[- {3}^{2} + 15 \left(3\right)\right]$

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

${\int}_{P}^{Q} \left(- 2 x + 15\right) \mathrm{dx} = \left(\frac{81}{4}\right)$

${\int}_{O}^{Q} f \left(x\right) \mathrm{dx} = {\int}_{O}^{P} \left({x}^{2}\right) \mathrm{dx} + {\int}_{P}^{Q} \left(- 2 x + 15\right) \mathrm{dx}$

$A = {\int}_{O}^{Q} f \left(x\right) \mathrm{dx} = 9 + \left(\frac{81}{4}\right)$

$A = {\int}_{O}^{Q} f \left(x\right) \mathrm{dx} = 9 + \left(\frac{81}{4}\right)$

$A = \left(\frac{36}{4}\right) + \left(\frac{81}{4}\right)$

$A = \left(\frac{117}{4}\right)$