USE CALCULUS TO FIND THE AREA A OF THE TRIANGLE WITH THE GIVEN VERTICES.

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Use calculus to find the location $A$ of the triangle with the provided vertices:$(0, 0), (4, 1), (1, 6)$

The formula for location is $A = int_a^bF(x) dx $


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HINT: It helps to sketch the region you"re taking care of. You can produce the equation of the lines in between $A = (0,0)$, $B = (1,6)$, and $C = (4,1)$.

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You understand that you deserve to uncover the location between two curves by doing $int_a^b( extTOP - extBOTTOM),dx$, which you"ll must separation this right into two integrals (why?). From there, you can find the area by just summing up the locations of the 2 resulting integrals.

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The three sides of that triangle are given by y= x/4 (from (0, 0) to (4, 1)), y= 6x (from (0, 0) to (1, 6)), y= (-3/5)x+ 33/5 (from (4, 1) to (1, 6)). y= x/4 is the lowest side so the location deserve to be calculated by integrating from that to the other 42 sides:$int_0^1 (6x- x/4)dx+ int_1^4 ((-3/5)x+ 33/5- x/4) dx$$int_0^1 23x/4 dx+ int_1^4 ((-17/20)x+ 33/5)dx$.

That have to be basic to combine.


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Plot the given points on a piece of graph paper and attach the dots. Then, you need to discover the functions for the lines that develop the triangle (they"re going to be linear functions). When you carry out all these actions, you will finish up with this integral:

$$int_0^1left(6x-frac14x ight),dx+int_1^4left<-frac53x+frac233-left(-frac14x ight) ight>,dx.$$

The numerical worth you"ll obtain resolving this integral is going to be equal to the location of that triangle.


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