Finding the area ! so hard

y = e^x and y = x - 1 , -2 ≤ x ≤ 0

Thank youuuuuuu

y = e^x is always above the y = x-1, so you can just subtract and then integrate

Int. [-2, 0] of (e^x - (x - 1) ) dx
Int. [-2, 0] of (e^x - x + 1) dx
= e^x - 1/2x^2 + x | [-2, 0]
= 1 - [e^ (-2) - 4 ]
Area = -e^(-2) - 3

Since you used integrals, you get NET signed area (which can be positive OR negative)
in this case, since most of the region lies BELOW the x-axis, the NET SIGNED area is negative

I'll guide you in the right direction. First what you need to do is graph the two equations. Once you have done that mark in the limits. Use integration on both equations, remembering that when you integrate e, you get the same thing. so you'll have e^x and x^2/2 - x. The area between these two curves is found by subtracting the smaller area under a curve from the larger one. (after you substitute in your limits in the integrated expressions). Use your graph for this.

You first have to graph both and see which curve is on top. Then find the definite intigral from -2 to 0 of the upper curve minus the lower curve.

I cant really draw it out because I have no idea how but I'll include a link which explains how to do these problems. The equation should end up being (e^x) - [(1/2)x^2 -x] after the first step.

Hope this helps and good luck!

I presume you mean the area enclosed by the two graphs.
You should sketch the graphs to see what is required.
The straight line is below the curve
so integrate (e^x - (x-1)dx =(e^x-x+1)dx from x=-2 to x=0.
You should get 5-e^(-2)

Dear
you have to sketch both functions and then integrate the expression
if you was not able to do so , pleas let me know
i will provide one online graphing calculator to make things easier
http://graph.seriesmathstudy.com/