The odd natural number a, such that the area of the region bounded by \(y = 1, y = 3, x = 0, x = y^a\) is \(\frac {364}{3}\), is equal to
- 3
- 5
- 7
- 9
The Correct Option is B
Solution and Explanation
\(|∫_1^3y^ady|=\frac {364}{3}\)
\(|\frac {1}{a+1}(y^{a+1})|_1^3=\frac {364}{3}\)
\(\frac {3a+1−1}{a+1}=±\frac {364}{3}\)
Solving with (+) sign,
\(\frac {3a+1−1}{a+1}=\frac {364}{3}\)
\(a=5\)
Solving with (-) sign,
\(\frac {3a+1−1}{a+1}=-\frac {364}{3}\)
No a exist
\(∴a=5\)
So, the correct option is (B): \(5\)
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Concepts Used:
Area under Simple Curves
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