My teacher (AP Calculus BC) gave me this problem:
"Let R be the shaded region in the first quadrant enclosed by the graphs of $y =e^{-x^2}$, $y=1-\cos x$, and the y-axis."
Obviously, this calls for a definite integral, which I wrote like this:
$$\int_0^b \left(e^{-x^2}-1+\cos x \right)dx$$
I put $b$ as the upper limit, as I hadn't found out yet.
However, this is my first question (elementary, I know): How would I find $b$?
I set up my equation like this:
$$e^{-x^2}=1-\cos x$$
I don't really know how to go on from here. I tried multiplying both sides by $1+\cos x$ but this didn't help.
Later, I used a solver to find the numerical approximation to the above equation, which gave me the upper bound of $0.942$.
Now, this leads to my second question. I have a definite integral: $$\int_0^{0.942} \left(e^{-x^2}-1+\cos x\right) dx$$
How would I evaluate this?
My first instinct was to split the integral up. When I split the integral, I ended up with this:
$$\int_0^{0.942}e^{-x^2}dx – \int_0^{0.942}1dx + \int_0^{0.942}\cos xdx$$
The second and third terms are very simple and easy to integrate. However, the first term, $\int_0^{0.942}e^{-x^2}dx$ would have to be solved with the error function. I only know the error function because of Math.SE, but is there any other way to solve this? Using the error function, $\int_0^{0.942}e^{-x^2}dx$ can be simplified to $\frac {\sqrt \pi} 2 \text{erf}(x)|_0^{0.942}$. This is supposed to be an AP Calculus BC problem, however, so I'm not sure if $\text{erf(x)}$ is the only way to go.
Best Answer
If you do not want to use the error function, expand the integrand as a Taylor series around $x=0$. This would give $$e^{-x^2}-1+\cos (x)=1-\frac{3 x^2}{2}+\frac{13 x^4}{24}-\frac{121 x^6}{720}+O\left(x^8\right)$$
Integrate termwise and use the bounds. This would give for the definite integral a value of $0.588605$ instead of $0.590962$ (relative error of $0.4$%).