Problem 27 - Calculus Early Transcendentals

Question

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R}(y+1) e^{x(y+1)} d A ; R=\\{(x, y): 0 \leq x \leq 1,-1 \leq y \leq 1\\}$$

Step-by-Step Solution

Answer

Question: Determine the best order of integration for evaluating the double integral $$\int_{-1}^{1}\int_{0}^{1}(y+1)e^{x(y+1)}dxdy$$ and find its value. Answer: The best order of integration is to proceed with the given order. The value of the double integral is $$e^2 - 3$$.

Step 1: Integrand simplification

The integrand is \((y+1)e^{x(y+1)}\). There is no simplification possible before integrating.

Step 2: Determine the best order of integration

The bounds are \(0 \leq x \leq 1\) and \(-1 \leq y \leq 1\). We have two possible orders of integration: $$\int_{-1}^{1}\int_{0}^{1}(y+1)e^{x(y+1)}dxdy$$ or $$\int_{0}^{1}\int_{-1}^{1}(y+1)e^{x(y+1)}dydx$$. Looking at the integrand, we don't have any clear advantage, so we will proceed with the given order.

Step 3: Integrate with respect to x

$$\int_{0}^{1}(y+1)e^{x(y+1)}dx$$ To integrate with respect to x, we perform a substitution: let \(u = (y+1)x\), so \(du = (y+1)dx\). Now the integral becomes $$\int (y+1)e^{(y+1)x}d(u/(y+1))= \int e^u du$$. The bounds of \(x\) are now \(0\) and \((y+1)\): $$\int_{0}^{y+1} e^u du$$ Now integrate to get $$E(u) = e^u$$ Evaluate at the bounds: $$E(y+1) - E(0) = e^{y + 1} - e^0 = e^{y + 1} - 1$$

Step 4: Integrate with respect to y

Now we integrate the result with respect to y: $$\int_{-1}^1 (e^{y + 1} - 1) dy$$ We split the integral into two parts: $$\int_{-1}^1 e^{y + 1} dy - \int_{-1}^1 dy$$ First integral: $$E'(y) = e^{y + 1}$$ Evaluate at the bounds: $$E'(1) - E'(-1) = e^{2} - e^0 = e^2 - 1$$ Second integral: $$Y(y) = y$$ Evaluate at the bounds: $$Y(1) - Y(-1) = 1 - (-1) = 2$$ Now we add the two results: $$\boxed{e^2 - 1 - 2 = e^2 - 3}$$