How to solve this problem step by step with application of integration? Mar 14, 2018

$\text{Area} = 48$

Explanation:

First, let's find the $x$-coordinates of the points where both lines cross the $x$-axis:

For $y = x + 8$: $y = 0 R i g h t a r r o w x + 8 = 0 \therefore x = - 8$

For $y = 8 - 2 x$: $y = 0 R i g h t a r r o w 8 - 2 x = 0 R i g h t a r r o w 2 x = 8 \therefore x = 4$

In order to calculate the requested area, our definite integral must then be evaluated in the interval $\left[- 8 , 4\right]$.

We can do this by halving the interval; by adding the area under $y = x + 8$ (from $- 8$ to $0$) to the area under $y = 8 - 2 x$ (from $0$ to $4$).

First, let's evaluate the indefinite integrals for each line:

$R i g h t a r r o w \int$ $\left(x + 8\right)$ $\mathrm{dx} = \frac{1}{2} {x}^{2} + 8 x$

and

$\int$ $\left(8 - 2 x\right)$ $\mathrm{dx} = 8 x - {x}^{2}$

Then, let's evaluate these integrals at the bounds mentioned earlier:

$R i g h t a r r o w | \frac{1}{2} {x}^{2} + 8 x {|}_{- 8}^{0} = | \left(\frac{1}{2} {\left(0\right)}^{2} + 8 \left(0\right)\right) - \left(\frac{1}{2} {\left(- 8\right)}^{2} + 8 \left(- 8\right)\right) | = | - \left(\frac{64}{2} - 64\right) | = | - \left(32 - 64\right) | = | - 32 | = 32$

and

$R i g h t a r r o w | 8 x - {x}^{2} {|}_{0}^{4} = | \left(8 \left(4\right) - {\left(4\right)}^{2}\right) - \left(8 \left(0\right) - {\left(0\right)}^{2}\right) | = | \left(32 - 16\right) - \left(0 - 0\right) | = | 16 | = 16$

Finally, the area is the sum of these two values:

$\therefore \text{Area} = 32 + 16 = 48$

Mar 14, 2018

The answer is $16$ , or so I'm pretty sure

Explanation:

So I'm gonna assume you don't know anything about integration.
The region is bounded by $y = 8 - 2 x$ , $y = x + 8$ and $y = 0$. It'll be easier if we look at these as two triangles divided by the Y-axis.

First you need to find where the two lines cut the X-axis, you do that by replacing the $y$ with $0$ , and so you get $0 = 8 - 2 x$ and $0 = x + 8$ and you get the $x$ from there, which are $- 8$ and $4$. These two numbers are the location of the vertices of the triangles.
$A = {\int}_{-} {8}^{0} \left(8 - 2 x\right) \mathrm{dx} + {\int}_{0}^{4} \left(x + 8\right) \mathrm{dx}$
$A = 16$

I really do hope i don't have to go any further then this, but if necessary i will help, I'm just having a bit of trouble with writing out the formulas.

Mar 14, 2018

$48$ square units.

Explanation: We can calculate the given area by finding Area A and area B and adding these to find total area.

To find area A we need the upper and lower bounds. We can see where these are from the graph, but it is always best to find them algebraically, graphs are not normally accurate enough.

These can be found by solving:

$x + 8 = 0 \implies x = - 8$

Our upper bound is at the y axis, $x = 0$

So we need the integral:

${\int}_{- 8}^{0} \left(x + 8\right) \mathrm{dx} = {\left[\frac{1}{2} {x}^{2} + 8 x\right]}_{- 8}^{0}$

Area$= {\left[\frac{1}{2} {x}^{2} + 8 x\right]}^{0} - {\left[\frac{1}{2} {x}^{2} + 8 x\right]}_{- 8}$

Plugging in upper and lower bounds:

Area$= {\left[\frac{1}{2} {\left(0\right)}^{2} + 8 \left(0\right)\right]}^{0} - {\left[\frac{1}{2} {\left(- 8\right)}^{2} + 8 \left(- 8\right)\right]}_{- 8}$

Area$= {\left[0\right]}^{0} - {\left[- 32\right]}_{- 8} = 32$ square units.

For area B

The upper bound is:

$8 - 2 x = 0 \implies x = 4$

So we need integral:

${\int}_{0}^{4} \left(8 - 2 x\right) \mathrm{dx} = {\left[8 x - {x}^{2}\right]}_{0}^{4}$

Area$= {\left[8 x - {x}^{2}\right]}^{4} - {\left[8 x - {x}^{2}\right]}_{0}$

Plugging in upper and lower bounds:

Area$= {\left[8 \left(4\right) - {\left(4\right)}^{2}\right]}^{4} - {\left[8 \left(0\right) - {\left(0\right)}^{2}\right]}_{0}$

Area$= {\left[8 \left(4\right) - {\left(4\right)}^{2}\right]}^{4} - {\left[8 \left(0\right) - {\left(0\right)}^{2}\right]}_{0} = 16$ square units.

Total area:

$A + B = 32 + 16 = 48$ square units.