(I6) Integration By Parts (Definite)#

By the end of the lesson you will be able to:

  • calculate a definite integral using integration by parts.

Lecture Videos#

The Notation#

When calculating a definite integral using integration by parts, we have two slightly different ways we can handle the evaluation (all of the integration-by-parts work will be exactly the same).

We can either evaluate each part of the formula separately:

\[ \int_a^b f(x)\cdot g'(x) \; dx = f(x)\cdot g(x) \biggr\rvert_{a}^{b} - \int_a^b g(x)\cdot f'(x)\; dx \]

or we can just wait until we have fully finished finding the antiderivative and then evaluate everything all at once:

\[ \int_a^b f(x)\cdot g'(x) \; dx = \biggr[f(x)\cdot g(x) - \int g(x)\cdot f'(x)\; dx \biggr]_{a}^{b} \]

Example 1#

Integrate the following:

\[ \int_{-1}^2 \dfrac{2x+5}{e^{3x}} \, dx \]
Solution (Click to see the steps.)

Before we start any integration, it will be helpful to rewrite this function as a product:

\[\int_{-1}^2 \dfrac{2x+5}{e^{3x}} \, dx = \int_{-1}^2 (2x+5)\cdot e^{-3x} \, dx\]

We’re going to use integration by parts, so our first step is to choose \(u\) and \(dv\):

Our choice for \(u\)
\[u=2x+5\]
Our choice for \(dv\)
\[dv = e^{-3x} \; dx\]

Next, we calculate \(du\) and \(v\) by differentiating and integrating respectively:

Differentiate
\[\begin{split}u &=2x+5 \\ du &= 2\cdot dx\end{split}\]
Integrate
\[\begin{split}dv &=e^{-3x} \; dx \\ v &=-\tfrac{1}{3}e^{-3x}\end{split}\]

Now we take the pieces we just calculated and plug them into the integration by parts formula.

The Formula
\[\int u \; dv = uv -\int v \; du\]

Our Pieces

\[\begin{split}u &=2x+5 \\ du &= 2\cdot dx\end{split}\]
\[\begin{split}dv &=e^{-3x} \; dx \\ v &=-\tfrac{1}{3}e^{-3x}\end{split}\]

Doing this we get:

\[\int_{-1}^2 (2x+5)\cdot e^{-3x} \, dx = \biggr[ -\tfrac{1}{3}(2x+5)e^{-3x}+\tfrac{2}{3}\int e^{-3x}\, dx\biggr]_{x=-1}^{x=2}\]

Finish calculating any remaining integrals:

\[\begin{split}&= \biggr[ -\tfrac{1}{3}(2x+5)e^{-3x}+\tfrac{2}{3}\int e^{-3x}\, dx\biggr]_{x=-1}^{x=2} \\ &= \biggr[ -\tfrac{1}{3}(2x+5)e^{-3x}+\tfrac{2}{3}\left(\tfrac{-1}{3}e^{-3x}\right)\biggr]_{x=-1}^{x=2}\end{split}\]

Finally, simplify the antiderivative as much as you desire and then evaluate.

\[\begin{split}&= \biggr[ -\tfrac{1}{3}e^{-3x}\left(2x+5+\tfrac{2}{3}\right)\biggr]_{x=-1}^{x=2} \\ &= \biggr( -\tfrac{1}{3}e^{-6}\left(4+5+\tfrac{2}{3}\right) \biggr) -\biggr( -\tfrac{1}{3}e^{3}\left(-2+5+\tfrac{2}{3}\right) \biggr)\end{split}\]

(Remember we plug the top number in first, the bottom number in second, and then we subtract.)

Example 2#

Integrate the following:

\[ \int_0^1 \tan^{-1}x \, dx \]
Solution (Click to see the steps.)

We’re trying to integrate an inverse function, so it might be helpful to remember back to our trick on how we integrated \(\ln x\). There, we rewrote the function as a product, and that’s what we’re going to do here as well:

\[ \left(\tan^{-1} x\right)\cdot 1 \]

So that we have the integral:

\[ \int_0^1 \left(\tan^{-1} x\right)\cdot 1 \cdot \, dx \]

Now we’re ready to choose \(u\) and \(dv\) by classifing the two parts of our function and remembering LIATE.

A for Algebraic
\[1\]
I for Inverse Trigonometric
\[\tan^{-1} x\]

Since we have an algebraic part and an inverse trigonometric part, the algebraic part is our choice for \(dv\) and the inverse trigonometric part is our choice for \(u\).

Our choice for \(u\)
\[u=\tan^{-1} x\]
Our choice for \(dv\)
\[dv = 1\cdot dx\]

Next, we calculate \(du\) and \(v\) by differentiating and integrating respectively:

Differentiate
\[\begin{split}u &=\tan^{-1} x \\ du &= \dfrac{1}{1+x^2}\cdot dx\end{split}\]
Integrate
\[\begin{split}dv &=1 \cdot dx \\ v &=x\end{split}\]

Now we take the pieces we just calculated and plug them into the integration by parts formula.

The Formula
\[\int u \; dv = uv -\int v \; du\]

Our Pieces

\[\begin{split}u &=\tan^{-1} x \\ du &= \dfrac{1}{1+x^2}\cdot dx\end{split}\]
\[\begin{split}dv &=1 \cdot dx \\ v &=x\end{split}\]

Doing this we get:

\[\begin{split}\int_0^1 \tan^{-1} x \,& dx = uv \biggr|_0^1 -\int_0^1 v \; du \\ &= \left(\tan^{-1} x\right)\cdot x \biggr|_0^1 -\int_0^1 x\cdot \dfrac{1}{1+x^2} \; dx\end{split}\]

We’re eventually going to finish integrating, but since this will require substitution rule, let’s evaluate that first piece to get it out of the way.

\[\begin{split}&= x\tan^{-1} x \biggr|_0^1 -\int_0^1 x\cdot \dfrac{1}{1+x^2} \; dx\\ &= \left(1\cdot \tan^{-1} 1\right)-\left(0\cdot \tan^{-1} 0\right) -\int_0^1 x\cdot \dfrac{1}{1+x^2} \; dx \\ &= \dfrac{\pi}{4} -\int_0^1 x\cdot \dfrac{1}{1+x^2} \; dx \end{split}\]

(Since \(\tan^{-1} 1 = \pi/4\)).

Now, to calculate this remaining integral:

\[\int_0^1 \dfrac{1}{1+x^2} \cdot x \; dx\]

We are going to do a \(u\)-substitution with:

\[\begin{split}u &= 1+x^2 \\ du &= 2x \cdot dx \longrightarrow \tfrac{1}{2}\cdot du = x dx \end{split}\]

And converting the limits of integration gives us the integral:

\[\int_0^1 \dfrac{1}{1+x^2} \cdot x \; dx = \int_1^2 \dfrac{1}{u} \cdot \dfrac{1}{2} \; du\]

Next we fnd the antiderivative and evaluate to get:

\[\begin{split}\int_1^2 \dfrac{1}{u} \cdot \dfrac{1}{2} \; du &= \tfrac{1}{2}\ln|u|\biggr|_{u=1}^{u=2}\\ &= \tfrac{1}{2}\ln|2|-\tfrac{1}{2}\ln|1|\\ &= \tfrac{1}{2}\ln 2\end{split}\]

Now, put it altogether to get our answer:

\[\begin{split}\int_0^1 \tan^{-1} x \, dx & = \dfrac{\pi}{4} -\int_0^1 x\cdot \dfrac{1}{1+x^2} \; dx \\ &=\tfrac{\pi}{4} -\tfrac{1}{2}\ln 2\end{split}\]