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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


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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