Problem 18 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{0}^{1} \int_{0}^{2 x} 15 x y^{2} d y d x$$

Step-by-Step Solution

Answer

Based on the given problem, the double integral of the function \(f(x, y) = 15xy^2\) over the triangular region in the xy-plane is found to be 8.

Step 1: Inner Integral Antiderivative

First, let's find the antiderivative of the inner integral with respect to \(y\). Since \(x\) is being treated as a constant in this case, we'll just integrate \(15xy^2\) with respect to \(y\): $$\int 15xy^2 dy = 15x\int y^2 dy = 15x\left[\frac{1}{3}y^3\right] = 5xy^3$$

Step 2: Evaluate Inner Integral

Now we'll compute the inner integral with respect to \(y\) by applying the limits of integration, \(0\) and \(2x\): $$\int_{0}^{2x} 15xy^2 dy = 5x[(2x)^3 - 0^3] = 5x(8x^3) = 40x^4$$

Step 3: Outer Integral Antiderivative

Now that we have the result of the inner integral, let's find the antiderivative of the outer integral with respect to \(x\): $$\int 40x^4 dx = 40\int x^4 dx = 40\left[\frac{1}{5}x^5\right] = 8x^5$$

Step 4: Evaluate Outer Integral

Finally, we'll compute the outer integral with respect to \(x\) by applying the limits of integration, \(0\) and \(1\): $$\int_{0}^{1} 40x^4 dx = 8[1^5 - 0^5] = 8(1) = 8$$ Thus, the value of the double integral is: $$\int_{0}^{1} \int_{0}^{2 x} 15 x y^{2} d y d x = 8$$