Question

Evaluate the definite integral. $$ \int_{0}^{2} \frac{x}{\sqrt{1+2 x^{2}}} d x $$

Short answer

The solution to the given definite integral is \(\left[\frac{1}{2}\sqrt{1+8}\right] - \left[\frac{1}{2}\sqrt{1+0}\right] = 1.5 - 0.5 = 1\).

Step-by-step solution

Identify the form of the integral

The integral is in the form \(\int \frac{x}{\sqrt{1+2 x^{2}}} dx\), we can see that the derivative of \(1 + 2x^2\) is \(4x\). This suggests we should use the substitution rule \(u = 1 + 2x^2\). This will convert the integral to a form that is easier to evaluate.

Apply substitution

Implement the substitution \(u = 1 + 2x^2\), then we need to find \(du\) in terms of \(dx\). By taking the derivative of \(u\), we get \(du = 4xdx\). However, we need \(dx\), so rearrange the equation to get \(dx = du/4x\). Now substitute \(u\) and \(dx\) in original integral and it becomes \(\int \frac{x}{\sqrt{u}} \cdot \frac{du}{4x}\)

Simplify the integral

After substitution, the integral simplifies to \(\frac{1}{4}\int \frac{1}{\sqrt{u}} du\). The integral now becomes simpler to solve.

Solve the integral

The antiderivative of \(\frac{1}{\sqrt{u}}\) is \(2\sqrt{u}\). Therefore, the antiderivative of entire integral \(\frac{1}{4}{\int \frac{1}{\sqrt{u}} du}\) is \(\frac{1}{2} \sqrt{u}\).

Back substitute and apply limits

Replace \(u\) by \(1+2x^{2}\), so we get \(\frac{1}{2} \sqrt{1+2x^{2}}\) as our antiderivative. To find the definite integral, evaluate this antiderivative at the upper and lower limits of the integral (0 and 2). So, the solution to the given definite integral is \(\left[\frac{1}{2}\sqrt{1+2(2^2)} \right] - \left[\frac{1}{2}\sqrt{1+2(0^2)}\right]\)

Key concepts

Substitution Rule

The substitution rule is a powerful technique used to simplify the process of evaluating integrals. It involves changing the variable of integration to make the integral simpler. In our exercise, we identified the expression \(1 + 2x^2\) as a suitable point for substitution. By letting \(u = 1 + 2x^2\), we are able to reframe the integral in terms of \(u\). This transforms the integral into a form that is often easier to manage. We replace \(dx\) with \(du\) based on the derivative \(du = 4x \, dx\). This step changes the integral into one with respect to \(u\), allowing us to tackle it using simpler methods. The key to success with substitution is correctly identifying the part of the integrand to replace, and ensuring all aspects of \(dx\) are accounted for when expressing \(du\). Essentially, substitution is like converting a complex path into a well-paved road, making it easier to reach our destination of finding the integral's value.

Antiderivative

Finding an antiderivative is an essential part of solving integrals. The antiderivative is a function whose derivative is the original function you started with. In simpler terms, it's about "undoing" differentiation. In our example, after applying substitution, we ended up with the integral of \(\frac{1}{\sqrt{u}}\) with respect to \(u\). The antiderivative of \(\frac{1}{\sqrt{u}}\) is \(2\sqrt{u}\), making this integral much easier to comprehend and calculate.By substituting back the original expressions, in this case, replacing \(u\) by \(1 + 2x^2\), we can express the antiderivative in terms of the original variable, \(x\). This antiderivative plays a crucial part when we later apply the integral limits to evaluate the definite integral. The key takeaway is recognizing the form of the integrand and knowing the antiderivative rules that help in finding a solution.

Evaluating Integrals

Evaluating integrals, particularly definite integrals, involves finding the antiderivative and applying the limits of integration. Think of it as finding the area under a curve within the specified interval. Once we obtain the antiderivative, the next step is to evaluate it at the given limits of the integral. This involves calculating the antiderivative at the upper limit and subtracting the value of the antiderivative at the lower limit. For our specific integral \(\int_{0}^{2} \frac{x}{\sqrt{1+2x^{2}}} dx\), we found the antiderivative to be \(\frac{1}{2} \sqrt{1+2x^2}\). Evaluate this antiderivative at the limits \(x = 2\) and \(x = 0\), subtract the two values, and you've successfully solved the definite integral.The final evaluation gives the result for the integral over the specified range, offering a precise measure of the total area under the function from those points.

Integral Limits

Integral limits define the range over which we are finding the integral. They are displayed as numbers at each bound of the integral sign in a definite integral expression.In definite integrals, these limits play an essential role as they define the interval used to evaluate the antiderivative. Definite integrals differ from indefinite integrals, as they result in numerical values rather than functions.For the integral \(\int_{0}^{2} \frac{x}{\sqrt{1+2x^{2}}} dx\), the limits are 0 and 2. These numbers represent the lower and upper bounds, respectively. After finding the antiderivative, we replaced \(x\) with these boundaries and solved for the difference. The limits are crucial as they constrain the calculation to a particular range, thus offering a comprehensive answer, such as total area or accumulated quantity, that reflects the behavior of the function within those bounds.