Question

In Exercises \(1-10\), evaluate the integral. $$ \int_{0}^{\sqrt{4-x^{2}}} x^{2} y d y $$

Short answer

The result of the integral is \( \frac{1}{2}x^{2}(4-x^{2})\).

Step-by-step solution

Identify the Integral to be evaluated

The integral to be evaluated is \(\int_{0}^{\sqrt{4-x^{2}}} x^{2} y dy\). Remember that when a variable (in this case \(y\)) is indicated in differential form (\(dy\)), it represents the variable of integration. So this is an integral with respect to \(y\).

Find the Antiderivative

To find the antiderivative, integrate the function \(x^{2}y\) with respect to \(y\). In this integral, the \(x^2\) term can be treated as constant since we are integrating with respect to \(y\). So the antiderivative is \(\frac{1}{2}x^{2}y^{2}\).

Apply the Limits of Integration

Now, apply the limits of the integral \(\sqrt{4-x^{2}}\) and \(0\) to the antiderivative \(\frac{1}{2}x^{2}y^{2}\). This gives us \(\frac{1}{2}x^{2}(\sqrt{4-x^{2}})^{2} - \frac{1}{2}x^{2}(0)^{2}\).

Simplify the Result

Simplify this to get the final result. This simplifies to \( \frac{1}{2}x^{2}(4-x^{2})\).

Key concepts

Definite Integral

The definite integral is a fundamental concept in calculus that measures the area under the curve of a function between two points on the x-axis, known as the limits of integration. When we evaluate the definite integral, we are essentially summing up infinitely small slices of area to find the total area enclosed by the graph of the function and the axis.

Using the given exercise, \[ \int_{0}^{\sqrt{4-x^{2}}} x^{2} y dy \], as an example, we're looking for the area under the curve of \( x^{2} y \) as \( y \) varies from 0 to \( \sqrt{4-x^{2}} \). This specific type of integral is known as an iterated integral, as it represents the area in a two-dimensional space where one variable is a function of another. This approach allows us to integrate more complex functions that involve multiple variables.

Antiderivative

In integral calculus, an antiderivative is a function whose derivative is the original function that we are integrating. It's known as the indefinite integral because it includes a constant of integration that represents an infinite number of possible antiderivatives.

In our exercise, the antiderivative of \( x^{2}y \) with respect to \( y \) is found by treating \( x^{2} \) as a constant, giving us \( \frac{1}{2}x^{2}y^{2} \) plus a constant. However, since we are dealing with a definite integral, this constant of integration cancels out when evaluating the limits of integration, leading us to the final result without needing to include it explicitly in the calculation.

Integration Limits

Integration limits define the boundaries over which the definite integral is calculated. They are essential because they determine the specific section of the graph we are interested in when finding the area under the curve.

In the exercise, the limits are from \( y = 0 \) to \( y = \sqrt{4-x^{2}} \). These tell us the starting and ending points of the integration process. By substituting these limits into the antiderivative, we calculate the definite integral’s value. We are essentially plugging the upper limit into the antiderivative, subtracting the result of the lower limit plugged into the antiderivative. This process leads us to the precise area value within the specified limits.