Problem 25 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{0}^{1} \int_{0}^{x} 2 e^{x^{2}} d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral $$\iint_{R} 2e^{x^2} \, dy\,dx$$, where R is the region defined by $$0 \leq y \leq x$$ and $$0 \leq x \leq 1$$. Answer: $$\frac{1}{2} (e - 1)$$

Step 1: Integrate with respect to y

First, integrate the given expression with respect to y over the interval $$0 \leq y \leq x$$. To accomplish this, treat e^{x^2} as a constant since it doesn't depend on y. $$\int_{0}^{x} 2 e^{x^{2}} d y = 2 e^{x^{2}} \int_{0}^{x} d y$$ Now, integrate with respect to y: $$2 e^{x^{2}} \left[ y \right]_0^x = 2 e^{x^{2}}(x - 0) = 2xe^{x^{2}}$$

Step 2: Integrate with respect to x

Next, integrate the result of the previous step with respect to x over the interval $$0 \leq x \leq 1$$: $$\int_{0}^{1} 2xe^{x^2} d x$$ To solve this integral, we will use the substitution method. Let $$u = x^2$$, so $$\frac{du}{dx} = 2x$$ and $$dx = \frac{du}{2x}$$. We should also change the limits of integration accordingly, so when $$x=0$$, $$u=0$$ and when $$x=1$$, $$u=1$$: $$\int_{0}^{1} e^u \frac{du}{2} = \frac{1}{2} \int_{0}^{1} e^u du$$ Now integrate this simpler expression with respect to u: $$\frac{1}{2} \left[ e^u \right]_0^1 = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)$$ So, the value of the given double integral is: $$\frac{1}{2} (e - 1)$$.