Integration of definite integrals using substitution

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Say we want to find

$I = \int\limits_1^3 {x{e^{3{x^2} + 4}}dx}$

We can here see that we can use the method of substitution since the derivative of the inner function 3x²+4 is x, except for a constant factor. We thus have

$\left| {\begin{array}{*{20}{c}} {u = 3{x^2} + 4} \\ {\frac{{du}}{{dx}} = 6x} \\ {du = 6x{\kern 1pt} dx} \\ {\frac{{du}}{6} = x{\kern 1pt} dx} \end{array}} \right.$

Thus gives us

$\begin{gathered} I = \int\limits_1^3 {x{e^{3{x^2} + 4}}dx} = \int\limits_1^3 {{e^{3{x^2} + 4}}xdx} \hfill \\ \quad = \int\limits_{x = 1}^3 {{e^u}\frac{{du}}{6}} = \frac{1}{6}\int\limits_{x = 1}^3 {{e^u}du} = \hfill \\ \quad = \frac{1}{6}\left[ {{e^u}} \right]_{x = 1}^3 = \frac{1}{6}\left[ {{e^{3{x^2} + 4}}} \right]_{x = 1}^3 \hfill \\ \quad = \frac{1}{6}\left( {{e^{3 \cdot {3^2} + 4}} - {e^{3 \cdot {1^2} + 4}}} \right) = \frac{1}{6}({e^{31}} - {e^7}) \hfill \end{gathered}$

Here we need to indicate that we still, in the end, will integrate over the interval x=1 to x=3 even if our variable, for the moment, is x. This we can do by writing x=1 as our lower integration limit.

We could also have calculated the new integration limits using the substitution u=3x²+4.

I.e.

$\left| \begin{gathered} u(1) = 3 \cdot {1^2} + 4 = 7 \hfill \\ u(3) = 3 \cdot {3^2} + 4 = 31\quad \quad \quad \hfill \end{gathered} \right.$

$\begin{gathered} I = \int\limits_1^3 {x{e^{3{x^2} + 4}}dx} = \int\limits_1^3 {{e^{3{x^2} + 4}}xdx} \hfill \\ \quad = \int\limits_7^{31} {{e^u}\frac{{du}}{6}} = \frac{1}{6}\int\limits_7^{31} {{e^u}du} = \hfill \\ \quad = \frac{1}{6}\left[ {{e^u}} \right]_7^{31} = \frac{1}{6}\left[ {{e^{3{x^2} + 4}}} \right]_7^{31} \hfill \\ \quad = \frac{1}{6}({e^{31}} - {e^7}) \hfill \end{gathered}$

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Last modified: Dec 28, 2023 @ 14:30