Question3). Evaluate by changing of the ordrof integration. (a) $\int_{0}^{2} \int_{1}^{e^{x}} d y d x$ (b) $\int_{0}^{1} \int_{x^{2}}^{2-x} x y d x d y$ (c) $\int_{0}^{4 a} \int_{x / 4 a}^{2 \sqrt{a x}} d y d x$.

Studdy Solution
Evaluate the integral $\int_{0}^{4a} \left[ x \right]_{4ay}^{\frac{y^2}{4a}} d y$:
$$= \int_{0}^{4a} \left( \frac{y^2}{4a} - 4ay \right) d y$$
$$= \int_{0}^{4a} \left( \frac{y^2}{4a} - 4ay \right) d y$$
Evaluate each term separately:
$$\int_{0}^{4a} \frac{y^2}{4a} \, dy = \frac{1}{4a} \left[ \frac{y^3}{3} \right]_{0}^{4a} = \frac{1}{4a} \cdot \frac{(4a)^3}{3} = \frac{64a^2}{3}$$
$$\int_{0}^{4a} -4ay \, dy = -4a \left[ \frac{y^2}{2} \right]_{0}^{4a} = -4a \cdot \frac{(4a)^2}{2} = -32a^3$$
Combine the results:
$$\frac{64a^2}{3} - 32a^3$$
The solutions to the integrals are: (a) $e^2 - 1$ (b) $\frac{3}{8}$ (c) $\frac{64a^2}{3} - 32a^3$