Definite Integral – Finding area between 3 lines – Exercise 7020

Exercise

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

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

Final Answer

Show final answer

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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