The odd natural number a, such that the area of the region bounded by y=1, y=3, x=0, x=y^a…

Question

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

Accepted Answer

$$\mathrm{a}$$ is a odd natural number and

$$\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}$$

$$ \Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}$$

$$ \Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}$$

Solving with $( - )$ sign,

$${{{3^{a + 1}} - 1} \over {a + 1}} = {{364} \over 3} \Rightarrow (a = 5)$$

Solving with $( + )$ sign,

$${{{3^{a + 1}} - 1} \over {a + 1}} = {{ - 364} \over 3}$$, No a exist

$$\therefore$$ $(a = 5)$

Solutions · 1