# The whole process, once again

• 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 whole process, once again, continued

• 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 $\lim_{n\to\infty}R_n=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x=\lim_{n\to\infty}\sum_{i=1}^n\frac{i^3}{n^3}\frac{1}{n}=\frac14.$

# The general setup

• 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 $$0\le x\le1$$.

# The distance problem

• 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.

# A first example of a definite integral

• 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.

# Definite integrals and signed areas

• 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.
• 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

# Properties of the definite integral

• 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