$\int_{- 1}^{2} \sqrt{5 x + 6} d x$

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Solution

Let $5 x + 6 = t$ , then on differentiating the equation we get, $5 d x = d t$ .
When $x = - 1$ then $t = 5 \times - 1 + 6 = 1$ and $x = 2$ then $t = 5 \times 2 + 6 = 16$
Thus,
$\int_{- 1}^{5} \sqrt{5 x + 6} d x = \frac{1}{5} \int_{1}^{16} \sqrt{t} d t = \frac{1}{5} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{t^{\frac{3}{2}}}{\frac{3}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}_{1}^{16} = \frac{2}{15} ⎡ ⎢ ⎣ 16^{\frac{3}{2}} - 1 ⎤ ⎥ ⎦ = \frac{2}{15} \times 63 = \frac{42}{5} = 8 \frac{2}{5}$

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