Question

What is the result of the integration \(∫_3^4∫_1^2(x^2+y)dxdy\)?

(a) \(\frac{83}{6} \)

(b) \(\frac{83}{3} \)

(c) \(\frac{82}{6} \)

(d) \(\frac{81}{6} \)

I got this question by my school teacher while I was bunking the class.

I'm obligated to ask this question of Change of Order of Integration: Double Integral topic in portion Multiple Integrals of Engineering Mathematics

Answer 1

Right answer is (a) \(\frac{83}{6} \)

The explanation is: Given: \(∫_3^4∫_1^2(x^2+y)dxdy\)

Integrating with respect to y first, we get,

\(∫_3^4(x^2(y)_1^2+(\frac{y^2}{2})_1^2)dx= ∫_3^4(x^2+\frac{3}{2}) dx\)

Next integrating with respect to x, we get,

\((\frac{x^3}{3})_3^4+\frac{3}{2}(x)_3^4=  \frac{37}{3}+\frac{3}{2}=\frac{83}{6}\)