##### Definite integral of the area

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Use a definite integral to find the area bounded by the graphs of the indicated equations over the given interval. y=2x+3; y=x-1; 6<x<8.

Aug 7th, 2014

We will take this integral for x from the constants 6 to 8

we need to find the function we will integrate. in general for eqns such as this the formula will be upper graph minus lower graph. This makes sense because if we take the area of the upper graph we have some extra space below the lower graph. We can then subtracts this extra area by taking the integral of the lower graph and subtracting it from the upper;

The upper graph can be found by finding the y value for x=6 for both equations

y=6-1=5

y=2*6+3=15

these are both lines, so the y=2x+3 will always be above y=x-1. If the were not lines and the interesected in the region between 6 and 8, we would have to make a complicated integral, considering this.

integral from x=6 to x=8 of (2x+3-x+1) dx

integral from x=6 to x=8 of (x+4) dx

= x^2/2 + 4x evaluated at x=6 and x=8

=32+32 - 18 - 24

=22

Aug 7th, 2014

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Aug 7th, 2014
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Aug 7th, 2014
Dec 7th, 2016
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