How do i find the shaded area of a curve using integration from this question (question in the picture) 

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

well, when we substitute x=2 in the gradient (dy/dx) we get e^-2 - 2 e^-2

which can be simplified to e^-2 (1-2) = -e^-2

hope that clarifies.

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

that is the equation of the tangent in the form, (y-y1) = m (x-x1)

where m is the calculated gradient and (x1,y1) is point of intersection of the curve and the tangent.

hope this helps.

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kimmiek | Student, Grade 11 | (Level 1) Honors

Posted on

Oh, one more question, for  e^-2 -2e^-2=-e^-2

The (e^-2 -2e^-2)  is from substituting x=2 right? 

How about the -e^-2? 

gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

Given y= `xe^-x`

and the point of intersection of the tangent is (`2,2/e^2` )

The shaded area is bound by the tangent and the curve between points x =0 and x=2.

First we need to find the equation of the tangent.

The gradient of tangent = dy/dx = `e^-x -xe^-x`

(using the product rule).

And at point x=2, gradient = e^-2 -2e^-2 = -e^-2

Using simple mathematics, equation of the tangent is given by,

y-2/e^2 = -e^-2 (x-2) 

on simplification, we get, y = -xe^-2 +4e^-2

Therefore the shaded area = area bound by tangent - area bound by the curve (between x=0 and x-2) 

= `int (-xe^-2 +4e^-2 - xe^-x) dx = -x^2/(2e^2) +4x/e^2 + xe^-x +e^-x`

between x=0 and x =2

substituting these limits of x, we get shaded area= 9/e^2 -1 = 0.218 square units.

hope this helps.

` `

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kimmiek's profile pic

kimmiek | Student, Grade 11 | (Level 1) Honors

Posted on

What's y-2/e^2=-e^-2(x-2)?  

How did you get it? 

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