Question

Evaluate the following integral.

∫(1 to 4)∫(0 to 4) (6x−3y)dxdy=

Answer 1

The double integral is evaluated as,

$\int_{1}^{4}\int_{0}^{4}(6x-3y)dxdy$ $=\int_{1}^{4}[6(\frac{x^2}{2})-3yx]_{0}^{4}dy$

$=\int_{1}^{4}[3(4^2-0^2)-3y(4-0)]dy$

$=\int_{1}^{4}(48-12y)dy$

$=[48y-12(\frac{y^2}{2})]_{1}^{4}$

$=[48(4-1)-6(4^2-1^2)]$

$=[48(3)-6(16-1)]$

$=144-6(15)$

$=144-90$

$=54$

Therefore, the double integral is $\int_{1}^{4}\int_{0}^{4}(6x-3y)dxdy=54$ .