If int_1^4 x sqrt{x^2-1} , dx = alpha(k)^beta | vedclass

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Question

(B) Let $I = \int_1^4 x \sqrt{x^2-1} \, dx$. Substitute $t = x^2 - 1$,so $dt = 2x \, dx$ or $x \, dx = \frac{dt}{2}$. When $x = 1$,$t = 1^2 - 1 = 0$.

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Let $I = \int_1^4 x \sqrt{x^2-1} \, dx$. Substitute $t = x^2 - 1$,so $dt = 2x \, dx$ or $x \, dx = \frac{dt}{2}$. When $x = 1$,$t = 1^2 - 1 = 0$. When $x = 4$,$t = 4^2 - 1 = 15$. Thus,$I = \int_0^{15} \sqrt{t} \frac{dt}{2} = \frac{1}{2} \int_0^{15} t^{1/2} \, dt$. $I = \frac{1}{2} \left[ \frac{t^{3/2}}{3/2} ight]_0^{15} = \frac{1}{2} 	imes \frac{2}{3} [t^{3/2}]_0^{15} = \frac{1}{3} (15)^{3/2}$. Comparing this with $\alpha(k)^\beta$,we get $\alpha = \frac{1}{3}$,$k = 15$,and $\beta = \frac{3}{2}$. Therefore,$\alpha \beta = \frac{1}{3} 	imes \frac{3}{2} = \frac{1}{2}$.

If $\int_1^4 x \sqrt{x^2-1} \, dx = \alpha(k)^\beta$,then $\alpha \beta$ is equal to

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