# Integration - Examples

On this page we display several examples to illustrate the concepts and theorems discussed in the Integration theme.

## Computing an integral using the definition...

**Problem**

Find
\[I = \int_0^3 e^{3x}dx\]
using the definition of the integral.

**Solution**

Consider the partition of \([0,3]\) with partition points \(P_n = \{0,\frac1n, \frac2n, \ldots, 3\}\). Then, since the integrand is increasing, we can use the values at the left endpoints as the heights for the lower rectangles, and the values at the right endpoints for the upper triangles.

\[U(f,P_n) = \sum_{i=1}^{3n} \frac 1n e^{3 \cdot \frac{i}{n}} = \frac1n \frac{e^9-1}{1-e^{-3/n}}. \] The last equality was obtained by using the formula for the sum of a geometric series. Moreover, we have that \[L(f,P_n) = \sum_{i=1}^{3n} \frac 1n e^{3 \cdot \frac{i-1}{n}} = \frac{U(f,P_n)}{e^{3/n}}.\] Therefore we get that \(\lim_{n \to \infty} L(f,P_n) = \lim_{n \to \infty} U(f,P_n)\), so there can only be one number between the lower and the upper sums. Therefore the integral exists, and has to be equal to the common value of these two limits. In other words,

\[\begin{align} I &= \lim_{n \to \infty} U(f,P_n) \\ &= (e^9-1) \lim_{n \to \infty}\frac{1/n}{1-e^{-3/n}} \\ &= (e^9-1) \lim_{n \to \infty}\frac{-1/n^2}{-\frac{3}{n^2}e^{-3/n}} \qquad (\mathrm{by \ l'Hôpital}) \\ &= \frac{e^9-1}{3}. \end{align}\]

## ...and using the Fundamental Theorem of Calculus

**Problem**

Find
\[I = \int_0^3 e^{3x}dx\]
using the Fundamental Theorem of Calculus.

**Solution**

Since \(g(x) = \frac13 e^{3x}\) is an antiderivative of the integrand, we get
\[I = g(3)-g(0) = \frac{e^9-1}{3}.\]

## Computing a definite integral using substitution

**Problem**

Compute the definite integral

\[\int_0^5 xe^{x^2}\mathrm{d}x.\]

**Solution**

We wish to find a function of \(x\), \(g(x)\), that we may substitute with \(u\) to simplify the expression into something that we can readily calculate the antiderivative of. Since \(u=g(x)\) implies \(\mathrm{d}u=g'(x)\mathrm{d}x\), we always end up dividing by the derivative of \(g(x)\). Hence it is essential that \(g'(x)\) itself is a part of the original expression (disregarding constants); we do not wish our integral to be dependent on both \(u\) and \(x\), for then we have complicated and not simplified the problem.

In our case then, the choice should be apparent: \(u=x^2\) yields \(\mathrm{d}x=\frac{\mathrm{d}u}{2x}\). Taking the constant \(1/2\) outside, we then have

\[\frac{1}{2}\int e^{u}\mathrm{d}u\]

But now one should be careful with the interval of integration; our original variable \(x\) went from \(0\) to \(5\), but our new variable \(u\) does not! \(x=0 \Rightarrow u=x^2=0\) and \(x=5 \Rightarrow u=x^2=25\). Thus we have reformulated the problem to

\[\frac{1}{2}\int_0^{25} e^{u}\mathrm{d}u,\]

which you can easily solve.

**Remark:** Instead of changing the limits, one may find an antiderivative of the new integrand (the antiderivative is \(1/2e^u\) in our case) and substitute back to our original variable to find an antiderivative of the original expression (i.e. \(1/2e^{x^2}\)).

## Computing an indefinite integral using substitution

**Problem**

Compute the indefinite integral
\[ I = \int \frac{dt}{t (\ln t)^3}. \]

**Solution**

Let \(u = \ln t\). Then \(du = \frac1t dt\), so we get
\[I = \int \frac{du}{u^3} = -\frac{1}{2u^2} + C = -\frac{1}{2(\ln t)^2}+C,\]
where \(C\) is an arbitrary constant.

## Computing the area of a plane region

**Problem:**

Calculate the area bounded by \(\mathrm{e}^x\) and \(6-5(x-1)^2\), to the left of \(x=1.5\).

**Solution:**

First we need to find the intersection point(s) to the left of \(x=1.5\). Since \(\mathrm{e}^x\) and its derivative are strictly increasing and positive on the whole real line and \(6-5(x-1)^2\) is a parabola opening downwards, they will have at most two intersection points.
By inspection, \(x=0\) is an intersection point, and the derivative of \(6-5(x-1)^2\) is greater than the derivative of \(\mathrm{e}^x\) in this point. Thus \(\mathrm{e}^x\) lies below \(6-5(x-1)^2\) to the immediate right of \(x=0\). Since \(\mathrm{e}^x\) grows faster than any polynomial, the second intersection point has to lie to the right of \(x=0\). Thus \(x=0\) is the left endpoint of our area.

By use of a calculator we find that \(\mathrm{e}^{1.5}<6-5(1.5-)^2\), and thus, by the mean-value theorem and the remarks above, there are no intersections in the interval \((0,1.5]\), and \(\mathrm{e}^x\) lies below \(6-5(x-1)^2\) on this interval. The area can then be expressed as

\[A=\int_0^{1.5}6-5(x-1)^2-\mathrm{e}^x \,dx.\] Using the Fundamental Theorem of calculus, we find \[A=[6x-\frac{5}{3}(x-1)^3-\mathrm{e}^x]_0^{1.5}=\frac{211}{24}-\mathrm{e}^{1.5}-\frac{2}{3}.\]