Finding the area bounded by two graphs

[Guest]

I've spent quite a bit of time on these two problems but I can't seem to get the correct answer. I would truly appreciate any help I can get.

1. y= 1/(x^2+1), y=0, x=-1, x=1

I've done the following for this problem:

a=1 and u=x therefore the integral is: arctanx from -1 to 1. So pi/4 - 3pi/4= -pi/2

I know that area can't be negative and the answer at the back of the book says pi/2. What am I doing wrong?

2. y=e^x, y=e^2, x=0

I've done the following for this problem:

The integral of e^x=e^x (right?) and the integral of e^2=e^2 (right?)

So I get: [e^2 -e^x] from 0 to 2 where I get, when I plugg them in and do b-a,
-e^2+1. Again the back of the book says that the actual answer is e^2+1. Why do I keep getting negative values? Please help me.

Thanks in advance!
Carlos

 

[soroban]

Hello, Carlos!

Your work is almost correct . . .

\(\displaystyle \L1)\;\,y\:=\:\frac{1}{x^2+1}\;\;y\,=\,0\;\;x\,=\,-1\;\;x\,=\,1\)

\(\displaystyle \L\int^{\;\;\;1}_{-1}\frac{dx}{x^2\,+\,1}\;=\;\arctan x\,\:|^{^1}_{_{-1}}\;=\;\arctan(1)\,-\,\arctan(-1)\)

. . \(\displaystyle \L=\;\frac{\pi}{4}\,-\,\left(-\frac{\pi}{4}\right)\;=\;\frac{\pi}{2}\)

\(\displaystyle \L2)\;y\,=\,e^x\;\;y\,=\,e^2 \;\;x\,=\,0\)

\(\displaystyle \L\int (e^2\,-\,e^x)\,dx \;=\;\left(e^2\cdot x\,-\,e^x\right)\,|^{^2}_{_0} \;=\; \left(e^2\cdot2\,-\,e^2\right)\,-\,\left(e^2\cdot0 \,-\,e^0\right)\)

. . \(\displaystyle \L=\;\left(2e^2\,-\,e^2\right)\,-\,(0\,-\,1) \;= \;e^2\,+\,1\)

 

[Guest]

Thanks a bunch soroban; I get what I was doing wrong now!