# Definite Integral – Finding area between 3 lines – Exercise 7020

Exercise

Find the area of the region bounded by the graphs of the equations:

$$y-4x=0, x+y=5, y=0$$

$$S=10$$

Solution

First, we find out how the area looks like:

$$y-4x=0\Longrightarrow y=4x$$

$$x+y=5\Longrightarrow y=-x+5$$

$$4x=-x+5$$

$$5x=5$$

$$x=1$$

$$y-4x=0$$

$$0-4x=0$$

$$4x=0$$

$$x=0$$

In the equation y=0 we set

$$x+y=5$$

and get

$$x+0=5$$

$$x=5$$

The area looks like this: Hence, it is a sum of two disjoint areas: We will calculate them seperately:

$$S=S_1+S_2$$

The first area:

$$S_1=\int_0^1 4x dx=$$

$$=[4\cdot\frac{x^2}{2}]_0^1=$$

$$=[2x^2]_0^1=$$

$$=2\cdot 1^2-2\cdot 0^2=$$

$$=2-0=$$

$$=2$$

The second area:

$$S_2=\int_1^5 -x+5 dx=$$

$$= [-\frac{x^2}{2}+5x]_1^5=$$

$$=-\frac{5^2}{2}+5\cdot 5-(-\frac{1^2}{2}+5\cdot 1)=$$

$$=-\frac{25}{2}+25+\frac{1}{2}-5=$$

$$=20-12\frac{1}{2}+\frac{1}{2}=$$

$$=8$$

Hence, we got

$$S_2=8$$

Lastly, we sum up the results:

$$S=S_1+S_2=$$

$$=2+8=10$$

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