- Let us find the area of the region bounded above by \(y=f(x)=e^x\), bounded below by \(y=g(x)=x\), and bounded on the sides by \(x=0\) and \(x=1\).
- The method to find this area is a generalization of the method to find the area under a graph.
- We'll calculate the area as a limit \(A=\lim_{n\to\infty}R_n\).
- In the partial sum \(R_n\), we once again split the interval \(I=[a,b]\) to \(n\) pieces of length \(\Delta x=\frac{b-a}{n}\).
- The difference is that in this case, the rectangles will have height \(f(x_i^*)-g(x_i^*)\).
- That is, we'll have \[ R_n=\sum_{i=1}^n(f(x_i^*)-g(x_i^*))\Delta x, \] and thus the area will be \[ A=\int_a^bf(x)-g(x)\,\mathrm dx=\int_0^1e^x-x\,\mathrm dx=e^x-\frac{x^2}{2}\Big|_0^1=e-\frac32. \]
- Click here for an illustration.

- Let us find the area of the region enclosed by the parabolas \(y=f(x)=x^2\) and \(y=g(x)=2x-x^2\).
- When the interval \([a,b]\) is not given, we have to find it. It will be the largest closed interval in which \(f(x)\ge g(x)\) all along or \(f(x)\le g(x)\) all along.
- For this, first we need to find the points \(x\) such that \(f(x)=g(x)\): \[\begin{align*} x^2=&2x-x^2\\ 2x^2=&2x, \end{align*}\] so the points we're looking for are \(x=0,1\).
- Then we need to decide which function has a larger value on the interval. For this, we substitute a point in the open interval \((0,1)\), for example \(0.5\): \[ f(0.5)=0.25<g(0.5)=0.75, \]
- This shows that we have \(g(x)\ge f(x)\) on \([0,1]\), therefore the area will be \[
A=\int_a^bg(x)-f(x)\,\mathrm dx=\int_0^12x-2x^2\,\mathrm dx=x^2-\frac23x^3\Big|_0^1=\frac13.
\]

- Click here for an illustration.

- Let us find the area bounded by the curves \(y=\sin x\), \(y=\cos x\), \(x=0\), and \(x=\frac{\pi}{2}\).
- Here, although the base interval \([a,b]=[0,\frac{\pi}{2}]\) is given, we still need to decide which function has a larger value.
- We need this because the definite integral calculates the net area. Therefore, over subintervals where \(f(x)\le g(x)\), the area between the curves gets subtracted, not added.
- Click here for an illustration. The positive areas are blue, and the negative ones are green.
- Therefore, again we need to find all points \(x\) in \([0,\frac{\pi}{2}]\) such that \(f(x)=g(x)\): \[ \sin x=\cos x \] when \(x=\frac{\pi}{4}\).
- Now we would need test points \(\frac{\pi}{6}\) from \((0,\frac{\pi}{4})\) and \(\frac{\pi}{3}\) from \((\frac{\pi}{4},\frac{\pi}{2})\), but alternatively we can just look at the graph or remember the shapes of the graphs of \(\sin x\) and \(\cos x\) to see that \(\cos x\ge \sin x\) on \([0,\frac{\pi}{4}]\), but \(\sin x\ge\cos x\) on \([\frac{\pi}{4},\frac{\pi}{2}]\).
- Therefore, the area is \[ A=\int_0^{\pi/4}\cos x-\sin x\,\mathrm dx+\int_{\pi/4}^{\pi/2}\sin x-\cos x\,\mathrm dx =[\sin x+\cos x]_0^{\pi/4}+[-\cos x-\sin x]_{\pi/4}^{\pi/2}=2\sqrt2-2. \]

- In general, we can say that the area between \(y=f(x)\) and \(y=g(x)\) over \([a,b]\) is given by \[
A=\int_a^b|f(x)-g(x)|\,\mathrm dx.
\]
- Exercises. 6.1: 14, 16, 18, 22, 28, 34, 46, 50, 56, 58