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### Section 5-8 : Substitution Rule for Definite Integrals

9. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

$\int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}}$

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Start Solution

The first step that we need to do is do the substitution.

At this point you should be fairly comfortable with substitutions. If you are not comfortable with substitutions you should go back to the substitution sections and work some problems there.

Before setting up the substitution we’ll need to break up the integral because each term requires a different substitution. Doing this gives,

$\int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} \,dt}} + \int_{{ - 2}}^{0}{{\frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}}$

The substitution for each integral is then,

$u = 3 + {t^2}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}v = 6t - 1$ Show Step 2

Here is the actual substitution work for this first integral.

\begin{align*}du & = 2t\,dt\hspace{0.25in}\,\, \to \hspace{0.25in}t\,dt = \frac{1}{2}du\\ t & = - 2:u = 7\hspace{0.75in}t = 0:u = 3\end{align*}

Here is the actual substitution work for the second integral.

\begin{align*}dv & = 6\,dt\hspace{0.25in}\,\, \to \hspace{0.25in}dt = \frac{1}{6}dv\\ t & = - 2:v = - 13\hspace{0.75in}t = 0:v = - 1\end{align*}

As we did in the notes for this section we are also going to convert the limits to $$u$$’s to avoid having to deal with the back substitution after doing the integral.

Here is the integral after the substitution.

$\int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \frac{1}{2}\int_{7}^{3}{{{u^{\frac{1}{2}}}\,du}} + \frac{3}{6}\int_{{ - 13}}^{{ - 1}}{{{v^{ - 2}}\,dv}}$ Show Step 3

The integral is then,

$\int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}} = \left. {\frac{1}{3}{u^{\frac{3}{2}}}} \right|_7^3 - \left. {\frac{1}{2}{v^{ - 1}}} \right|_{ - 13}^{ - 1} = \frac{1}{3}\left( {{3^{\frac{3}{2}}} - {7^{\frac{3}{2}}}} \right) - \frac{1}{2}\left( { - 1 - \left( { - \frac{1}{{13}}} \right)} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\left( {{3^{\frac{3}{2}}} - {7^{\frac{3}{2}}}} \right) + \frac{6}{{13}}}}$