Question

Evaluate the integral and subsequent iterated integral. (a) \(\int_{0}^{x}\left(\frac{1}{1+x^{2}}\right) d y\) (b) \(\int_{1}^{2} \int_{0}^{x}\left(\frac{1}{1+x^{2}}\right) d y d x\)

Short answer

\( \frac{1}{2} \ln \left( \frac{5}{2} \right) \).

Step-by-step solution

Evaluate the inner integral (a)

Given the integral \( \int_{0}^{x} \left( \frac{1}{1+x^{2}} \right) dy \). Since there is no \( y \) variable in the function \( \frac{1}{1+x^{2}} \), the integral becomes simply \( \frac{1}{1+x^{2}} \cdot y \) evaluated from 0 to \( x \). This results in \( \frac{x}{1+x^{2}} \).

Evaluate the iterated integral (b)

Now, given the iterated integral \( \int_{1}^{2} \int_{0}^{x} \left( \frac{1}{1+x^{2}} \right) dy \, dx \), using the result from Step 1, we substitute in \( \frac{x}{1+x^{2}} \) for the inner integral. Thus, the iterated integral is \( \int_{1}^{2} \frac{x}{1+x^{2}} \, dx \).

Simplify the outer integral expression

Solve the outer integral \( \int_{1}^{2} \frac{x}{1+x^{2}} \, dx \). Use the substitution \( u = 1 + x^2 \), resulting in \( du = 2x \, dx \) or \( \frac{du}{2} = x \, dx \). Adjust the limits for \( x = 1 \) to \( u = 2 \) and for \( x = 2 \) to \( u = 5 \). The integral becomes \( \frac{1}{2} \int_{2}^{5} \frac{1}{u} \, du \).

Solve the simplified integral

The integral \( \frac{1}{2} \int_{2}^{5} \frac{1}{u} \, du \) simplifies to \( \frac{1}{2} [\ln|u|]_{2}^{5} \). Evaluate this expression: \( \frac{1}{2} ( \ln 5 - \ln 2 ) = \frac{1}{2} \ln \left( \frac{5}{2} \right) \).

Finalize the answer

Thus, the value of the iterated integral \( \int_{1}^{2} \int_{0}^{x} \left( \frac{1}{1+x^{2}} \right) dy \, dx \) is \( \frac{1}{2} \ln \left( \frac{5}{2} \right) \).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals, focusing on the accumulation of quantities. It essentially deals with finding the total from differential parts. Integrals can be thought of as the reverse operation of derivatives.

In integral calculus, we encounter two fundamental types: definite and indefinite integrals.

• Definite integrals have bounds and provide specific values representing the area under a curve between two limits. For instance, when you see an integral with limits, such as \( \int_{a}^{b} f(x) \, dx \), it's definite.
• Indefinite integrals do not have specified bounds, thus representing a family of functions with a constant of integration. An example is \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant.

In iterated integrals, like the one in the problem, the integration process happens in steps. First, we integrate with respect to one variable, and then we use that result to perform another integration with respect to another variable. This method is crucial in higher dimensions and is widely applied in multivariable calculus.

Substitution Method

The substitution method is a powerful technique used to simplify the integration process. It's particularly useful when dealing with complex integrals. The goal is to transform the integral into a simpler form, often by changing the variable.

To use substitution:

• Select a substitution \( u \) that simplifies the integrand.
• Calculate \( du \), which involves finding the derivative of \( u \).
• Adjust the limits of integration to match the new variable \( u \) if dealing with definite integrals.
• Rewrite the integral in terms of \( u \) and solve it.
• Finally, substitute back the original variable if necessary.

In the provided example, we used \( u = 1 + x^2 \), which helped turn the integral into a simpler form, \( \frac{1}{2} \int \frac{1}{u} \, du \). This transformation made the rest of the integration more manageable by turning a rational function into something straightforward.

Logarithmic Integration

Logarithmic integration is a technique often used when the integrand involves rational functions that integrate to logarithms. This is a specific case in integration where the antiderivative involves the natural logarithm function.

Consider the integral \( \int \frac{1}{u} \, du \), which evaluates to \( \ln|u| + C \). This formula is key in solving integrals that appear in the form of \( \frac{1}{u} \).

In the step-by-step solution, after substitution, we arrived at \( \frac{1}{2} \int_{2}^{5} \frac{1}{u} \, du \). This integral directly fits the logarithmic integration form.

Evaluating, we have:

• Calculate \( \ln|u| \) over the specified limits \([2, 5]\).
• Subtract the logarithms \( [ \ln 5 - \ln 2 ] \).
• Use properties of logarithms to simplify, yielding \( \ln \left( \frac{5}{2} \right) \).

Finally, multiply by any constants outside the integral, giving the final answer as \( \frac{1}{2} \ln \left( \frac{5}{2} \right) \).