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Evaluate the integral using substitution

by RoRi

Use the method of substitution to evaluate the following integral:

    \[ \int_{-\frac{2}{3}}^{\frac{1}{3}} \frac{x \, dx}{\sqrt{2-3x}}. \]

Let

    \begin{align*} u &= 2 - 3x &\implies && x &= \frac{2-u}{3} \\ du &= -3 \, dx & \implies && dx &= -\frac{1}{3} \, du. \end{align*}

For the limits of integration under this substitution we have

    \begin{align*} u \left( -\frac{2}{3} \right) &= 4 \\ u \left( \frac{1}{3} \right) &= 1 \end{align*}

Then, we can evaluate the integral,

    \begin{align*} \int_{-\frac{2}{3}}^{\frac{1}{3}} \frac{x \, dx}{\sqrt{2 - 3x}} &= -\frac{1}{3} \int_{u \left(-\frac{2}{3} \right)}^{u \left( \frac{1}{3} \right)} \left( \frac{1}{3}(2 - u)u^{-\frac{1}{2}} \right) \, du \\ &= -\frac{1}{9} \int_4^1 \left( 2u^{-\frac{1}{2}} - u^{\frac{1}{2}}\right) \, du \\ &= -\frac{1}{9} \left. \left( 4u^{\frac{1}{2}} - \frac{2}{3}u^{\frac{3}{2}} \right) \right|_4^1 \\ &= -\frac{1}{9} \left( 4 - \frac{2}{3} - 8 + \frac{16}{3} \right) \\ &= -\frac{2}{27}. \end{align*}

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