Basic Examples on Definite Integrals Set 4

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Example –8

Determine a positive integer \(n \le 5\) such that

\[\int\limits_0^1 {{e^x}{{\left( {x - 1} \right)}^n}dx = 16 - 6e} \]

Solution: Since n will not turn to be a very large integer, one might be tempted to try out various values of n in the given relation, starting from 1 onwards, and see which one fits. This trial-and-error approach might quickly give a result in this particular example, but what would we do it n was possibly larger?

The generally followed approach in such examples, where the integral can be characterised by positive integer n (called the order of the integral), is to express it in terms of a lower order integral. If we denote the n th order integral by In , we should try to express I n  in terms of Ik  where k < n. Such a relation is called a recursive relation. We can then simply use this relation repeatedly on any order and obtain the integral of the next order (instead of every time repeating the calculation of integration again)

We will now use this approach on the current example:

Let \({I_n} = \int\limits_0^1 {{e^x}{{\left( {x - 1} \right)}^n}dx} \)

To simplify In , first of all let \(x - 1 = t\, \Rightarrow dx = dt\) and the limits become –1 to 0.

Thus,

\[\begin{align}&{I_n} = \int\limits_{ - 1}^0 {{e^{t + 1}}{t^n}dt} \\\,\,\,\,\,\, &\;\;\;= e\int\limits_{ - 1}^0 {{e^t} \cdot {t^n}\,\,dt}\end{align}\]

We now use integration by parts to solve this integral, taking t n as the first function:

\[\begin{align}&{I_n} = e\left\{ {\left. {{t^n} \cdot {e^t}} \right|_{ - 1}^0 - n\int\limits_{ - 1}^0 {\left( {{t^{n - 1}} \cdot \int {{e^t}dt} } \right)dt} } \right\}\\ &\quad= e\left\{ {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{e} - n\int\limits_{ - 1}^0 {{t^{n - 1}} \cdot {e^t}dt} } \right.\\&\quad= {\left( { - 1} \right)^{n + 1}} - n{I_{n - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\end{align}\]

We have thus established the relation between I n  and In– 1 in (1)

Now observe that I0 can easily be evaluated:

\[\begin{array}{l}{I_0} = e\int\limits_{ - 1}^0 {{e^t}dt} \\\,\,\,\,\,\, = e - 1 \end{array}\]

Using (1) repeatedly, we can now obtain all the higher order integrals:

\[\begin{array}{l}{I_1} = {\left( { - 1} \right)^{1 + 1}} - 1.{I_0} = 2 - e\\{I_2} = {\left( { - 1} \right)^{2 + 1}} - 2.{I_1} = - 5 + 2e\\{I_3} = {\left( { - 1} \right)^{3 + 1}} - 3.{I_2} = 16 - 6e\end{array}\]

n = 3 is therefore the positive integer we had set out to determine. Notice the power of the recursive relation that we obtained in (1). Using that relation, it was just a matter of minor calculations to successively determine \(I_1\) , \(I_2\) and \(I_3\) from \(I_0\) .  Without (1), we would have to apply integration by parts everytime, had we used the trial-and-error approach.

Lets look at another example of this sort.

Example –9

Evaluate \(\int\limits_0^\infty {{e^{ - x}}{x^n}dx} \)

Solution: Notice that no matter what n be,

\[\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - x}}{x^n}} \right) = 0\]

so that we will obtain a finite area under the curve.

Let \({I_n} = \int\limits_0^\infty {{e^{ - x}}{x^n}dx} \)

We apply integration by parts on In :

\[\begin{align}&\left. {{I_n} = - {x^n}{e^{ - x}}} \right|_0^\infty + n\int\limits_0^\infty {{x^{n - 1}}{e^{ - x}}dx} \\\,\,\,\,\, &\quad= 0 + n{I_{n - 1}}\end{align}\]

Thus, our recursive relation is

\[{I_n} = n{I_{n - 1}}\]

We use this repeatedly now:

\[\begin{array}{l}{I_n} = n{I_{n - 1}}\\\,\,\,\,\,\, = n\left( {n - 1} \right){I_{n - 2}}\\\,\,\,\,\,\, = n\left( {n - 1} \right)\left( {n - 2} \right){I_{n - 3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ddots \\\,\,\,\,\,\, = n!{I_0}\end{array}\]

\({I_0}\) is simple to determine:

\[\begin{align}&{I_0} = \int\limits_0^\infty {{e^{ - x}}dx} \\\,\,\,\,\,\,&\quad\left. { = - {e^{ - x}}} \right|_0^\infty \\\,\,\,\,\, &\quad= 1\end{align}\]

Thus, \({I_n} = n!\)