Question

Evaluate the double integral. $$ \int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) d x d y $$

Short answer

The value of the double integral is 56

Step-by-step solution

Integrate with respect to x

Integral of \(3x^2\) with respect to x is \(x^3\), integral of \(2y^2\) with respect to x is \(2y^2x\) (note that y is treated as a constant), integral of 1 with respect to x is x. Thus the integral becomes: \[\int_{1}^{2} [x^3 - 2y^2x + x]_{x=0}^{x=4} dy\] Evaluate the values of x at 0 and 4, subtracting the lower limit from the upper limit for each.

Evaluate x Integral

After substituting x=4 and x=0 into [x^3 - 2y^2x + x], we get: \[\int_{1}^{2} (64 - 8y^2 + 4) dy\] This results in \[\int_{1}^{2} (68 - 8y^2) dy\] Now we are ready to proceed to integrate with respect to y.

Integrate with respect to y

The integral of 68 with respect to y is 68y, and the integral of -8y^2 with respect to y is \(- \frac{8y^3}{3}\). Thus the integral becomes: \[[68y - \frac{8y^3}{3}]_{y=1}^{y=2}\] . Evaluate the values of y at 1 and 2, subtracting the lower limit from the upper limit for each.

Evaluate Final Result

After substituting y=2 and y=1 into [68y - \frac{8y^3}{3}], we get \(136 - \frac{64}{3}\) - \(68 - \frac{8}{3})\). Simplifying gives the final result.