- Let us review the entire process of calculating an area as a limit.
- This time, let's calculate the area under the curve \(y=f(x)=x^3\), for \(0\le x\le1\).
- The area will be a limit of the sum \(R_n\) as \(n\to\infty\). The sum \(R_n\) is called a
*partial sum*. Let's see how to build up that sum.

- First of all, we want to split up the interval \([0,1]\) to \(n\) parts of equal length. These interval parts are referred to as
*subintervals*.- Since we want \(n\) subintervals of equal length, each of them should have length \(\frac{1}{n}\). This length is denoted by \(\Delta x\).
- Let's fix the following \((n+1)\) endpoints for the subintervals. \[ x_0=0,\,x_1=\frac{1}{n},\,x_2=\frac{2}{n},\dotsc,x_{n-1}=\frac{n-1}{n},\,x_n=1. \]
- Since the partial sum \(R_n\) will be expressed with the Sigma notation \(\sum_{i=1}^n\), we need to refer to the \(i\)-th subinterval \(I_i\), as the indexing variable \(i\) traverses the sequence \(1,2,\dotsc,n-1,n\).
- The \(i\)-th subinterval will have endpoints \(x_{i-1}\) and \(x_i\). That is, we'll have \[ I_i=[x_{i-1},x_i]=[\frac{i-1}{n},\frac{i}{n}]. \]
- Note that indeed, every one of the \(I_i\) has length \[ x_i-x_{i-1}=\frac{i}{n}-\frac{i-1}{n}=\frac{1}{n}=\Delta x. \]

- The area is approximated by the sum of the areas of the rectangles, with sides \([x_{i-1},x_i]=[\frac{i-1}{n},\frac{i}{n}]\) and \([0,f(x_i^*)]\), where \(x_{i-1}\le x_i^*\le x_i\) is a point on the interval \(I_i\).
- There are three common choices for the
*sample point*\(x_i^*\):- They can be the right endpoints of the intervals, that is \(x_i^*=x_i=\frac{i}{n}\).
- They can also be the left endpoints, that is \(x_i^*=x_{i-1}=\frac{i-1}{n}\).
- Or they can be the midpoints, that is \(x_i^*=\frac{x_{i-1}+x_{i}}{2}=\frac{2i-1}{2n}\).
- Click here for a demonstration. As a fourth choice, I have included the random placement. You can see that for small \(n\), these all give different partial sums, but as \(n\) increases, they all seem to converge to \(\frac14\).
- Exercise. Prove that \[ \lim_{n\to\infty}R_n=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x=\lim_{n\to\infty}\sum_{i=1}^n\frac{i^3}{n^3}\frac{1}{n}=\frac14. \]

- Let's get back to the general setup, that is where \(f(x)\) is a nonnegative function on a closed interval \([a,b]\).
- Let's first define the partial sum \(R_n\).
- Once again, we want to split up the interval \([a,b]\) to \(n\) subintervals of equal length \(\Delta x=\frac{b-a}{n}\).
- To this end, we fix the \((n+1)\) endpoints \[ x_0=a,\,x_1=a+\Delta x,\,x_2=a+2\Delta x,\,\dotsc,\,x_{n-1}=a+(n-1)\Delta x,\,x_n=b. \]
- Then the subintervals will be \[ I_1=[x_0,x_1]=[a,a+\Delta x],\,I_2=[x_1,x_2]=[a+\Delta x,a+2\Delta x],\,\dotsc,\,I_n=[x_{n-1},x_n]=[a+(n-1)\Delta x,b]. \]
- In every one of the subintervals \(I_i=[x_{i-1},x_i]=[a+(i-1)\Delta x,a+i\Delta x]\), we choose a point \(x_{i-1}\le x_i^*\le x_i\).
- The partial sum \(R_n\) is the sum of the areas of the rectangles with sides \([x_{i-1},x_i]\) and \([0,f(x_i^*)]\).
*Theorem.*The limit \[ \lim_{n\to\infty}R_n=\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x \] exists, and it does not depend on the choice of the \(x_{i-1}\le x_i^*\le x_i\).*Definition.*We call the limit in the Theorem the*area*under \(y=f(x)\) for \(0\le x\le1\).

- Let's now consider the
*distance problem*.- Suppose that an object is performing rectilinear motion with velocity \(v(t)\ge0\).
- If \(v(t)\) is constant on an interval \([t_{i-1},t_i]\), then the distance travelled on that interval is \((t_i-t_{i-1})v(t_i)\).
- Therefore, we can approximate the distance by approximating the velocity with a piecewise constant function, and we get the same formula as for areas: \[ d=\lim_{n\to\infty}\sum_{i=1}^nv(t_i^*)\Delta t. \]

*Warning.*In the next section, we'll see that this method only calculates areas and distances for nonnegative functions. For general functions, it gives*signed areas*and*displacement*.

- Let's now consider the same formula \[
\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x
\] for the function \(f(x)=x^3\) over the interval \([-1,1]\).
- Looking at the approximations, the partial sums seem to converge to 0.
- If we choose the midpoints as sample points, we can see that the partial sums themselves are 0 if \(n\) is even.
- Exercise. Prove that using the midpoints as sample points, the partial sums are 0.
- The reason is that if you use the same formula for a function \(f(x)\) which is not nonnegative, then it will calculate the
*signed area*under the graph \(y=f(x)\) over the interval \([a,b]\). This means that the areas are added with a positive sign where \(f(x)\ge0\), and the areas are added with a negative sign where \(f(x)\le0\). - In this particular case, we get a 0, because the function is symmetrical with respect to the origin, and therefore the area of the negative part is precisely \((-1)\) times the area of the positive part.

*Definition.*Let \(f(x)\) be a function defined on the interval \([a,b]\).- If the limit \[
\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x
\] exists, it's called the
*definite integral*of the function \(f(x)\) on the interval \([a,b]\), and it is denoted by \[ \int_a^bf(x)\,\mathrm dx. \] In this case, we say that the function \(f(x)\) is*integrable*on \([a,b]\). - More precisely, these are called
*Riemann integrals*. They were invented by the German Mathematician Bernhard Riemann (1826-1866). By the way, his PhD advisor was Herr Gauß (he was a bit older by then). - Correspondingly, the partial sums \(R_n=\sum_{i=0}^nf(x_i^*)\Delta x\) are also called
*Riemann sums*. - Signed areas are also called
*net areas*. Another way to put that notion is that the net area is the difference \(A_+-A_-\), where \(A_+\) is the area of the region above the \(x\)-axis, and \(A_-\) is the area of the region below the \(x\)-axis.

- If the limit \[
\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x
\] exists, it's called the
*Theorem.*If \(f(x)\) is continuous on \([a,b]\), or it has only a finite number of jump discontinuities, then \(f(x)\) is integrable on \([a,b]\).- Exercise. 5.2.22

- We have the following formulas.
- Let \(c\) be any constant. Then we have \(\int_a^bc\,\mathrm dx=c(b-a)\).
- \(\int_a^bf(x)+g(x)\,\mathrm dx=\int_a^bf(x)\,\mathrm dx+\int_a^bg(x)\,\mathrm dx\).
- \(\int_a^bcf(x)\,\mathrm dx=c\int_a^bf(x)\,\mathrm dx\).
- \(\int_a^bf(x)\,\mathrm dx=\int_a^cf(x)\,\mathrm dx+\int_c^b\,\mathrm dx\).
- Exercise. Prove the first three in general, and the fourth one in case \(a<c<b\), and \(f(x)\ge0\) on \([a,b]\).
- Exercises. 5.2: 36, 38, 40.

- We also have the following comparison properties.
- Suppose that \(f(x)\ge0\) for \(a\le x\le b\). Then we have \(\int_a^bf(x)\,\mathrm dx\ge0\).
- Suppose that \(f(x)\ge g(x)\) for \(a\le x\le b\). Then we have \(\int_a^bf(x)\,\mathrm dx\ge\int_a^bg(x)\,\mathrm dx\).
- If \(m\le f(x)\le M\) for \(a\le x\le b\), then we have \(m(b-a)\le\int_a^bf(x)\,\mathrm dx\le M(b-a)\).
- Exercises. E.42, 5.2.70