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If $\int^4 x \sqrt{x^2-1} d x=\alpha(k)^\beta$, then $\alpha \beta$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}$
$\int_1^4 x \sqrt{x^2-1} d x=\alpha(k)^\beta$
Let $I=\int_1^4 x \sqrt{x^2-1} d x$
Let $x^2-1=t$
$\Rightarrow x d x=\frac{d t}{2}$
Upper limit $=t=4^2-1=15$
Lower limit $=t=1^2-1=0$
$\begin{aligned} I & =\int_0^{15} \sqrt{t} \frac{d t}{2} \\ I & =\frac{1}{2} \times \frac{2}{3}\left[t^{3 / 2}\right]_0^{15} \\ I & =\frac{1}{3}\left[(15)^{3 / 2}\right]=\alpha k^\beta \\ \Rightarrow \alpha & =\frac{1}{3}, k=15 \text { and } \beta=\frac{3}{2} \\ \therefore \alpha \beta & =\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\end{aligned}$
Let $I=\int_1^4 x \sqrt{x^2-1} d x$
Let $x^2-1=t$
$\Rightarrow x d x=\frac{d t}{2}$
Upper limit $=t=4^2-1=15$
Lower limit $=t=1^2-1=0$
$\begin{aligned} I & =\int_0^{15} \sqrt{t} \frac{d t}{2} \\ I & =\frac{1}{2} \times \frac{2}{3}\left[t^{3 / 2}\right]_0^{15} \\ I & =\frac{1}{3}\left[(15)^{3 / 2}\right]=\alpha k^\beta \\ \Rightarrow \alpha & =\frac{1}{3}, k=15 \text { and } \beta=\frac{3}{2} \\ \therefore \alpha \beta & =\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\end{aligned}$
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