Problem 9 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{1}^{3} \int_{0}^{\pi / 2} x \sin y d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the iterated integral \(\int_{1}^{3} \int_{0}^{\pi / 2} x \sin y d y d x\). Answer: The value of the iterated integral is \(-\frac{26}{3}\).

Step 1: Evaluate the inner integral with respect to \(y\)

$$ \int_{0}^{\pi/2} x \sin y dy $$We need to integrate this by parts, so let's assign:$$ u = x \implies du = dx \\ dv = \sin y dy \implies v = -\cos y $$Now, let's apply the integration by parts formula in the integral:\(\displaystyle\int u dv = uv - \int v du\). We get:$$ \int_{0}^{\pi/2} x \sin y dy = x(-\cos y)\Big|_0^{\pi/2} - \int_{0}^{\pi/2} (-\cos y) dx \\ = x(\cos y)\Big|_0^{\pi/2} -x\int_{0}^{\pi/2} (\cos y) dy $$Now, we need to evaluate the remaining integral with respect to \(y\). By integrating \(\cos y dy\) we get:$$ x(\cos y)\Big|_0^{\pi/2} -x\int_{0}^{\pi/2} (\cos y) dy = x(\cos y)\Big|_0^{\pi/2} -x(\sin y)\Big|_0^{\pi/2} $$After evaluating the integral at the limits, we have:$$ x(\cos y)\Big|_0^{\pi/2} -x(\sin y)\Big|_0^{\pi/2} = x(\cos(\pi/2) - \cos(0)) -x(\sin(\pi/2) - \sin(0)) \\ = x(0 - 1) - x(1 - 0) = -x^2 $$The inner integral becomes:$$ \int_{0}^{\pi/2} x \sin y dy = -x^2 $$Now, we'll proceed with the outer integral.

Step 2: Evaluate the outer integral with respect to \(x\)

$$ \int_{1}^{3} -x^2 dx $$Now we need to find the antiderivative of \(-x^2\). Let's calculate that:$$ \int -x^2 dx = -\frac{1}{3}x^3 $$Next, we'll apply the limits of integration to evaluate the integral:$$ -\frac{1}{3}x^3\Big|_1^3 = -\frac{1}{3}(27 - 1) = -\frac{1}{3}(26) =-\frac{26}{3} $$So, the value of the iterated integral is:$$ \int_{1}^{3} \int_{0}^{\pi / 2} x \sin y d y d x = -\frac{26}{3}