First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{4} 3^{\sqrt{2 x + 1}} d x$ is

Easy

A

$\frac{6}{\log 3} (13 - \frac{4}{\log 3})$
B

$\frac{66}{\log 3}$
C

$\frac{6}{\log 3} (13 - \frac{5}{\log 3})$
D

none of these

Solution

Putting $2 x + 1 = t^{2},$ we have $d x = t d t,$ so
$\begin{array}{l} \Rightarrow \int_{0}^{4} 3^{\sqrt{2 x + 1}} d x = \int_{1}^{3} 3^{t} t d t = {\frac{t \cdot 3^{t}}{\log 3}|}_{1}^{3} - \frac{1}{\log 3} \int_{1}^{3} 3^{t} d t \\ = \frac{(3) \cdot 3^{3} - 3}{\log 3} - \frac{1}{(\log 3)^{2}} [3^{3} - 3^{1}] = \frac{78}{\log 3} - \frac{24}{(\log 3)^{2}} \end{array}$

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