Question

how to find the area between two curves
y=e^x, y=x-1, -2<=x<=0

Answer 1

do you know calculus?
the "height" of the region is the top curve minus the lower curve
the width of a small rectangle is dx
integrate along x from -2 to 0: height * dx
see graph.

Answer 2

i found it on the calculator but want to know it alegbraic

Answer 3

what did you get for the area ?
it's not algebra, it is calculus that you use.

Answer 4

\[\int\limits_{-2}^{0}e ^{x} and \int\limits_{-2}^{0}x-1 \]

i got 4.8646 on the calculator but i want to know how to get the answer w/o one

Answer 5

the area is height * dx
where height is the top curve e^x minus the lower curve x-1
i.e. height = e^x - (x-1)
or height= e^x -x +1
and and the area is
\[ \int_{-2}^0 e^x -x +1 \ dx \]

Answer 6

You can "cheat" by noticing the area below the x-axis is a trapezoid (or a triangle on top of a rectangle), and has area 4
so we could say the total area is the area beneath the e^x curve plus the 4
\[ area = 4 + \int_{-2}^0 e^x \ dx \]
this should also give the same answer.

Answer 7

\[ \int_{-2}^0 e^x -x +1 \ dx \\ = e^x - \frac{x^2}{2} +x \ \bigg|_{-2}^0 \\
= (1 - 0 +0) - (e^{-2} -\frac{4}{2} -2) \\
= 5 - e^{-2}
\]
at this point, you need a calculator to evaluate e^-2
but if you don't have one, you can roughly estimate the expression:
\( e \approx 2.7\). If we call it 3, then 3^2 = 9, and 3^-2 is 1/9 or about .11
so (very roughly) the answer is 5-0.11 = 4.89