Question

Evaluate the following definite integrals. $$\int_{1 / \sqrt{3}}^{1} \frac{1}{x^{2} \sqrt{1+x^{2}}} d x$$

Short answer

The value of the definite integral is $$\frac{1}{2}\left[-\sqrt{2}+\sqrt{\frac{4}{3}}\right]$$

Step-by-step solution

Perform substitution

We will perform a substitution to simplify the integral. Let \(u = x^2\), so \(du = 2x\,dx\). We can rewrite the integral as follows: $$\int \frac{1}{u \sqrt{1+u}} \frac{1}{2} du$$ And now, we need to change the limits of the integral according to the substitution. When \(x = 1/\sqrt{3}\), we get \(u = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\), and when \(x = 1\), we get \(u = 1^2 = 1\). So, the integral now looks like this: $$\frac{1}{2}\int_{1/3}^{1} \frac{1}{u\sqrt{1+u}} du$$

Find the antiderivative

To find the antiderivative, we can use the following formula for the integral of a function \(f(u) = \frac{1}{u \sqrt{1 + u}}\). The antiderivative is given by: $$\int\frac{1}{u\sqrt{1+u}}du=-\sqrt{1+u} + C$$ where \(C\) is the constant of integration. Now we have the antiderivative, we can proceed to find the value of the definite integral.

Apply the Fundamental Theorem of Calculus

The definite integral is equal to the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit, according to the Fundamental Theorem of Calculus: $$\int_{1/3}^{1} \frac{1}{u\sqrt{1+u}} du = \left(-\sqrt{1+u}\right)\Big|_{1/3}^{1}$$

Evaluate the antiderivative at the limits

We will now evaluate the antiderivative at the limits \(1/3\) and \(1\): $$-\sqrt{1+u}\Big|_{1/3}^{1} = \left[-\sqrt{1+1} + \sqrt{1+\frac{1}{3}}\right] = \left[-\sqrt{2} + \sqrt{\frac{4}{3}}\right]$$

Answer the integral

From the calculations above, the value of the definite integral is: $$\int_{1/\sqrt{3}}^{1} \frac{1}{x^{2}\sqrt{1+x^2}} dx = \frac{1}{2}\left[-\sqrt{2}+\sqrt{\frac{4}{3}}\right]$$ Thus, the definite integral is equal to \(\frac{1}{2}\left[-\sqrt{2}+\sqrt{\frac{4}{3}}\right]\).

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of finding integrals. This method involves changing variables to make the integral easier to solve. In the context of the given exercise, we performed a substitution to tackle the integral with a complex expression.

• First, we selected a substitution: let \(u = x^2\).
• This meant that \(du = 2x \, dx\), which we rearranged to express \(dx\) in terms of \(du\).
• The substitution allowed us to rewrite the integral in terms of \(u\), transforming it from a function of \(x\) into a simpler integral that is easier to evaluate.

A critical part of the substitution process is changing the limits of the integral, which correspond to the new variable. This ensures that we accurately evaluate the definite integral:

• Lower limit: When \(x = \frac{1}{\sqrt{3}}\), \(u = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\).
• Upper limit: When \(x = 1\), \(u = 1^2 = 1\).

After substitution, the integral became \(\frac{1}{2} \int_{1/3}^{1} \frac{1}{u\sqrt{1+u}} \, du\). This new integral is simpler to work with and helps us find the solution more efficiently.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in calculus that connects differentiation and integration. It consists of two main parts, but in solving definite integrals, we focus on the second part, which provides a method to evaluate them:

• According to the theorem, if \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then the definite integral of \(f\) from \(a\) to \(b\) is given by \(F(b) - F(a)\).

This means that once we find an antiderivative, or an indefinite integral, we can evaluate the definite integral very straightforwardly:

• We calculate \(F(x)\) (i.e., the antiderivative) at the upper bound, which is \(b\).
• Subtract the antiderivative evaluated at the lower bound, \(a\), from the result at \(b\).

In the solution to the original problem, after determining the antiderivative, we applied the theorem to find:

• \(-\sqrt{1+u}\bigg|_{1/3}^{1}\) simplifies to \[-\sqrt{2} + \sqrt{\frac{4}{3}}\].
• Thus, using the limits, we were able to compute the definite integral's value.

This efficient process highlights the theorem's importance in resolving definite integrals swiftly once the antiderivative is in place.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. Finding an antiderivative is crucial when calculating definite integrals, as it forms the basis for the Fundamental Theorem of Calculus.

In the context of the problem, we found the antiderivative of the function \(\frac{1}{u\sqrt{1+u}}\). Here's how:

• We determined that an antiderivative of \(\frac{1}{u\sqrt{1+u}}\) is \(-\sqrt{1+u} + C\),

where \(C\) represents the constant of integration.

• Finding the antiderivative involved recognizing the format of the integral and knowing the standard results or rules of integration.
• These rules often involve logical deduction and application of known integral formulas to arrive at the antiderivative.

Antiderivatives remove derivatives' effects, allowing us to return to the original function, providing the solution needed for evaluating integrals.
Once an antiderivative is found, solving the definite integral becomes straightforward: evaluate the antiderivative at the specific limits and subtract, effectively solving the integral.