Question

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer. $$\int_{y=1}^{2} \int_{x=\sqrt{y}}^{y^{2}} x d x d y$$

Short answer

2.35

Step-by-step solution

Set up the Double Integral

Identify the limits of integration for both variables. The integral is given as \( \int_{y=1}^{2} \int_{x=\sqrt{y}}^{y^{2}} x \, dx \, dy \). The outer integral bounds are from 1 to 2 with respect to \( y \), and the inner integral bounds are from \( \sqrt{y} \) to \( y^{2} \) with respect to \( x \).

Integrate with Respect to x

Perform the inner integral first. Compute \( \int_{\sqrt{y}}^{y^{2}} x \, dx \). The antiderivative of \( x \) with respect to \( x \) is \( \frac{x^{2}}{2} \). Evaluate this from \( \sqrt{y} \) to \( y^{2} \): \[ \int_{\sqrt{y}}^{y^{2}} x \, dx = \left. \frac{x^{2}}{2} \right|_{\sqrt{y}}^{y^{2}}. \]

Evaluate the Antiderivative at the Bounds

First, evaluate the antiderivative at the upper bound \( y^{2} \): \[ \frac{(y^{2})^{2}}{2} = \frac{y^{4}}{2} \] Next, evaluate at the lower bound \( \sqrt{y} \): \[ \frac{(\sqrt{y})^{2}}{2} = \frac{y}{2} \] Subtract the lower bound's result from the upper bound's result: \[ \frac{y^{4}}{2} - \frac{y}{2} \] Thus, the result of the inner integral is \( \frac{y^{4}}{2} - \frac{y}{2} \).

Integrate with Respect to y

Now, perform the outer integral \( \int_{1}^{2} \left( \frac{y^{4}}{2} - \frac{y}{2} \right) \, dy \). First, separate the integral: \[ \int_{1}^{2} \frac{y^{4}}{2} \, dy - \int_{1}^{2} \frac{y}{2} \, dy \]

Integrate Each Term Separately

Integrate \( \frac{y^{4}}{2} \): \[ \int_{1}^{2} \frac{y^{4}}{2} \, dy = \frac{1}{2} \int_{1}^{2} y^{4} \, dy = \frac{1}{2} \left. \frac{y^{5}}{5} \right|_{1}^{2} = \frac{1}{10}(32 - 1) = \frac{31}{10} \]Integrate \( \frac{y}{2} \): \[ \int_{1}^{2} \frac{y}{2} \, dy = \frac{1}{2} \int_{1}^{2} y \, dy = \frac{1}{2} \left. \frac{y^{2}}{2} \right|_{1}^{2} = \frac{1}{4}(4 - 1) = \frac{3}{4} \]

Combine the Results

Subtract the result of the second integral from the first: \[ \frac{31}{10} - \frac{3}{4} = \frac{31}{10} - \frac{7.5}{10} = \frac{31-7.5}{10} = \frac{23.5}{10} = 2.35 \]

Key concepts

Limits of Integration

In a double integral, we need to define the bounds for both the inner and outer integrals.
The given double integral is written as:ewline \ \[\int_{y=1}^{2} \int_{x=\sqrt{y}}^{y^{2}} x \, dx \, dy\].
For the outer integral with respect to \(y\), the limits are from 1 to 2.
The inner integral with respect to \(x\) has limits that depend on the value of \(y\). Here, \(x\) varies from \(\sqrt{y}\) to \(y^{2}\).
Understanding these limits is crucial as they dictate the region over which the integration is performed.
Always identify these limits clearly to avoid errors in your calculations.
Recognizing the bounds helps in correctly setting up the integral and is the first crucial step.

Antiderivative

The antiderivative, or the indefinite integral, is crucial in solving integrals.
It represents the reverse process of differentiation.
For example, in the double integral from the exercise:
\ \[\int_{\sqrt{y}}^{y^{2}} x \, dx\], we need to find the antiderivative of \(x\) with respect to \(x\).
The antiderivative of \(x\) is \(\frac{x^2}{2}\). Next, evaluate this antiderivative at the given bounds:
At the upper bound \(y^2\):
\ \[\left. \frac{x^{2}}{2} \right|_{\sqrt{y}}^{y^{2}} = \frac{(y^{2})^{2}}{2} = \frac{y^{4}}{2}\]
At the lower bound \(\sqrt{y}\):
\ \[\frac{(\sqrt{y})^{2}}{2} = \frac{y}{2} \]
Subtracting the lower bound's evaluation from the upper bound's yields:
\ \[\frac{y^{4}}{2} - \frac{y}{2} \]
Antiderivatives simplify the integration process and are fundamental to solving both definite and indefinite integrals.

Definite Integral

A definite integral represents the accumulation of quantities over a specific interval.
It gives a numerical value rather than an expression.
In our example, after integrating with respect to \(x\), we perform the outer integral: \ \[\int_{1}^{2} \left( \frac{y^{4}}{2} - \frac{y}{2} \right) \, dy\].
This is broken down into two separate integrals:
\ \[\int_{1}^{2} \frac{y^{4}}{2} \, dy - \int_{1}^{2} \frac{y}{2} \, dy\].
Each of these integrals needs to be solved individually:
For the first term:
\ \[\frac{1}{2} \left. \frac{y^5}{5} \right|_{1}^{2} = \frac{1}{10}(32 - 1) = \frac{31}{10}\]
And for the second term:
\ \[\frac{1}{2} \left. \frac{y^2}{2} \right|_{1}^{2} = \frac{1}{4}(4 - 1) = \frac{3}{4}\]
Then, subtracting the results:
\ \[\frac{31}{10} - \frac{3}{4} = \frac{31 - 7.5}{10} = \frac{23.5}{10} = 2.35\].
The definite integral thus provides a specific, tangible value.
This process is fundamental in applications ranging from physics to engineering and economics.