The odd natural number a such that the area o | vedclass

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(B) The area $A$ of the region bounded by $y = 1, y = 3, x = 0,$ and $x = y^a$ is given by the integral with respect to $y$:$A = \int_{1}^{3} x \, dy

Vedclass · Accepted Answer

$5$

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(B) The area $A$ of the region bounded by $y = 1, y = 3, x = 0,$ and $x = y^a$ is given by the integral with respect to $y$:$A = \int_{1}^{3} x \, dy = \int_{1}^{3} y^a \, dy$Evaluating the integral:$A = \left[ \frac{y^{a+1}}{a+1} \right]_{1}^{3} = \frac{3^{a+1} - 1^{a+1}}{a+1} = \frac{3^{a+1} - 1}{a+1}$Given that $A = \frac{364}{3}$,we have:$\frac{3^{a+1} - 1}{a+1} = \frac{364}{3}$By testing odd natural numbers for $a$:If $a = 5$,then $a+1 = 6$:$\frac{3^6 - 1}{6} = \frac{729 - 1}{6} = \frac{728}{6} = \frac{364}{3}$Thus,the value of $a$ is $5$.

The odd natural number $a$ such that the area of the region bounded by $y = 1, y = 3, x = 0,$ and $x = y^a$ is $\frac{364}{3}$ is equal to:

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