Integral Evaluation Explained

This problem requires us to determine the value of a given double integral. Double integrals are used to integrate a function of two variables over a region in the xy-plane. They are often evaluated as iterated integrals, meaning we integrate with respect to one variable at a time. The expression we need to evaluate is:

\[\int^2_0\int^x_0y\ dy\ dx\]

To solve an iterated integral, the standard procedure is to work from the inside out. This means we first evaluate the innermost integral with respect to its variable, and then use that result to evaluate the outer integral.

Step-by-Step Integral Calculation

Inner Integral Evaluation: Integrating with respect to \(y\)

The first step is to evaluate the inner integral, which is \(\int^x_0y\ dy\). In this part, we treat \(x\) as a constant and integrate the function \(y\) with respect to \(y\).

• The antiderivative of \(y\) with respect to \(y\) is \(\frac{y^2}{2}\).
• Now, we apply the limits of integration, which are from \(y=0\) to \(y=x\): \[\int^x_0y\ dy = \left[\frac{y^2}{2}\right]^x_0\]
• Substitute the upper limit \(x\) and the lower limit 0 into the expression for \(y\): \[= \frac{(x)^2}{2} - \frac{(0)^2}{2}\] \[= \frac{x^2}{2} - 0\] \[= \frac{x^2}{2}\]

Thus, the result of the inner integral evaluation is \(\frac{x^2}{2}\). This result now becomes the integrand for our next step.

Outer Integral Evaluation: Integrating with respect to \(x\)

Now, we use the result from the inner integral, \(\frac{x^2}{2}\), as the function to be integrated in the outer integral. This integral is with respect to \(x\), from the lower limit of 0 to the upper limit of 2.

\[\int^2_0\frac{x^2}{2}\ dx\]

• We can factor out the constant \(\frac{1}{2}\) from the integral: \[= \frac{1}{2}\int^2_0x^2\ dx\]
• The antiderivative of \(x^2\) with respect to \(x\) is \(\frac{x^3}{3}\).
• Next, we apply the limits of integration, from \(x=0\) to \(x=2\): \[= \frac{1}{2}\left[\frac{x^3}{3}\right]^2_0\]
• Substitute the upper limit 2 and the lower limit 0 into the expression for \(x\): \[= \frac{1}{2}\left(\frac{(2)^3}{3} - \frac{(0)^3}{3}\right)\] \[= \frac{1}{2}\left(\frac{8}{3} - 0\right)\] \[= \frac{1}{2} \cdot \frac{8}{3}\] \[= \frac{8}{6}\]
• Finally, simplify the fraction to its lowest terms: \[= \frac{4}{3}\]

Final Value of the Double Integral

By completing both stages of the iterated integral calculation, we find that the value of the given double integral \(\int^2_0\int^x_0y\ dy\ dx\) is \(\frac{4}{3}\). This process is crucial for solving many problems in multivariable calculus, illustrating how to correctly handle definite integrals with multiple variables and varying limits.