Question

Sketch the region of integration and evaluate the following integrals as they are written. $$\int_{0}^{\pi / 2} \int_{y}^{\pi / 2} 6 \sin (2 x-3 y) d x d y$$

Short answer

Answer: The value of the double integral is 3.

Step-by-step solution

Identify the region of integration

We have a double integral with respect to x and y. The region of integration is a rectangle, where y ranges from 0 to \(\pi/2\), and for each value of y, x ranges from y to \(\pi/2\). This is a triangular region with vertices at \((0,0)\), \((\pi/2, 0)\), and \((\pi/2, \pi/2)\).

Evaluate the inner integral

First, we need to evaluate the integral with respect to x: $$\int_{y}^{\pi / 2} 6 \sin (2 x-3 y) d x$$ To solve this, we use the substitution method: Let \(u = 2x - 3y\), then \(du = 2dx\). We also need to change the limits of integration according to the new variable u. When \(x = y\), \(u = 2y - 3y = -y\). When \(x = \pi/2\), \(u = \pi-3y\). So the integral becomes: $$\int_{-y}^{\pi - 3y} 6 \sin u \cdot \frac{1}{2} du = 3 \int_{-y}^{\pi - 3y} \sin u \, du$$ Now integrate with respect to u: $$ 3[-\cos u \Big|_{-y}^{\pi - 3y}] = 3[-\cos(\pi - 3y) + cos(-y)] $$

Evaluate the outer integral

Now we can evaluate the integral with respect to y: $$\int_{0}^{\pi / 2} [3(-\cos(\pi - 3y) + \cos(-y))] dy$$ The integral can be broken down into two separate integrals: $$3\left[ \int_{0}^{\pi / 2} -\cos(\pi - 3y) dy + \int_{0}^{\pi / 2} \cos(-y) dy \right]$$ Using the substitution method for the first integral, let \(v = \pi - 3y\), then \(dv = -3 dy\). When \(y = 0\), \(v = \pi\). When \(y = \pi/2\), \(v = 0\). So the first integral becomes: $$\int_{\pi}^{0} -\frac{1}{3} \cos(v) dv = \frac{1}{3} \int_{0}^{\pi} \cos(v) dv$$ Now, we can compute the two integrals: $$3\left[\frac{1}{3} \int_{0}^{\pi} \cos(v) dv + \int_{0}^{\pi / 2} \cos(y) dy \right]$$ The antiderivatives for both are \(\sin\), so we have: $$3\left[\frac{1}{3} [\sin(v)\Big|_{0}^{\pi}] + [\sin(y)\Big|_{0}^{\pi / 2}] \right]$$ $$3\left[\frac{1}{3} [\sin(\pi) - \sin(0)] + [\sin(\pi / 2) - \sin(0)]\right]$$ Since \(\sin(\pi) = \sin(0) = 0\) and \(\sin(\pi / 2) = 1\), our final answer is: $$3\left[\frac{1}{3} [0] + [1]\right] = \boxed{3}$$

Key concepts

Region of Integration

When working with double integrals, the region of integration is crucial to determine. In our exercise, we have an integral in the form \[ \int_{0}^{\pi / 2} \int_{y}^{\pi/2} 6 \sin(2x-3y) \, dx \, dy \]Here, the limits of integration for \(y\) are from 0 to \(\pi/2\), and for each fixed \(y\), \(x\) ranges from \(y\) to \(\pi/2\). This defines a triangular region in the \(xy\)-plane.
To visualize it, the vertices of this triangle are at

• \((0, 0)\)
• \((\pi/2, 0)\)
• \((\pi/2, \pi/2)\)

Sketching this region can help you understand the problem better. It shows where the integration is performed and highlights the bounds that you'll work with when solving the integral.

Substitution Method

The substitution method for integrals simplifies the function you are integrating. For this exercise, we first need to evaluate the inner integral with respect to \(x\). This process involves substitution, where we let \(u = 2x - 3y\). - You'll find \(du = 2dx\), meaning \(dx = \frac{1}{2}du\).

An important step is to adjust the limits of integration to fit \(u\):

• When \(x = y\), \(u = -y\).
• When \(x = \pi/2\), \(u = \pi - 3y\).

This transforms the integral into: \[ 3 \int_{-y}^{\pi - 3y} \sin u \, du \]Applying the substitution makes the integral easier to evaluate and ensures that it can be further simplified. Always remember to adjust the limits accordingly to avoid errors during integration.

Trigonometric Integration

Trigonometric integrals involve functions like sine and cosine, which have specific integration rules. Here, after substitution, we simplify the integral:\[ 3 \int_{-y}^{\pi - 3y} \sin u \, du \]The antiderivative of \(\sin u\) is \(-\cos u\). Hence, the expression evaluates to:\[ 3[-\cos u \Big|_{-y}^{\pi - 3y}] \]In the next step, compute the difference:\[-3 \cos(\pi - 3y) + 3 \cos(-y)\]This result forms part of the outer integral \[ \int_{0}^{\pi / 2} [3 (-\cos(\pi - 3y) + \cos(-y))] \, dy \]Solve each integral individually by recognizing standard trigonometric integrals. Breaking down complex trigonometric forms into sums and differences helps simplify and compute these integrals more effectively.