Evaluate The Integral:$\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, Dx$

Introduction

Calculus is a branch of mathematics that deals with the study of continuous change, particularly in the context of functions and limits. One of the fundamental concepts in calculus is integration, which is used to find the area under curves and other mathematical functions. In this article, we will evaluate the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$, which is a classic example of an integral that can be solved using various techniques.

Background

The integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is a definite integral, which means that it has a specific upper and lower limit. In this case, the lower limit is $-1$ and the upper limit is $\sqrt{3}$. The integrand is $\frac{1}{1+x^2}$, which is a rational function. Rational functions are a type of function that can be expressed as the ratio of two polynomials.

Techniques for Evaluating the Integral

There are several techniques that can be used to evaluate the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$. Some of the most common techniques include:

• Substitution: This involves substituting a new variable into the integral, which can simplify the integrand and make it easier to evaluate.
• Integration by parts: This involves differentiating one function and integrating the other function, which can be used to evaluate integrals that involve products of functions.
• Partial fractions: This involves expressing a rational function as a sum of simpler fractions, which can be used to evaluate integrals that involve rational functions.

Evaluating the Integral using Substitution

One way to evaluate the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is to use substitution. Let's substitute $x = \tan(u)$, which implies that $dx = \sec^2(u) du$. We can then rewrite the integral as:

$$\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{1+\tan^2(u)} \sec^2(u) du$$

Simplifying the integrand, we get:

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{1+\tan^2(u)} \sec^2(u) du = \int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sec^2(u)} \sec^2(u) du$$

Evaluating the integral, we get:

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sec^2(u)} \sec^2(u) du = \left[ u \right]_{-\frac{\pi}{4}}^{\frac{\pi}{3}}$$

Simplifying the result, we get:

$$\left[ u \right]_{-\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$$

Evaluating the Integral using Integration by Parts

Another way to evaluate the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is to use integration by parts. Let's choose $u = \arctan(x)$ and $dv = dx$. We can then rewrite the integral as:

$$\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx = \int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx = \left[ \arctan(x) \right]_{-1}^{\sqrt{3}}$$

Evaluating the integral, we get:

$$\left[ \arctan(x) \right]_{-1}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(-1) = \frac{\pi}{3} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$$

Conclusion

In this article, we evaluated the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ using two different techniques: substitution and integration by parts. Both techniques resulted in the same answer, which is $\frac{7\pi}{12}$. This example illustrates the importance of having multiple techniques at one's disposal when evaluating integrals.

Final Answer

The final answer to the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is $\boxed{\frac{7\pi}{12}}$.

Introduction

In our previous article, we evaluated the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ using two different techniques: substitution and integration by parts. In this article, we will answer some common questions that readers may have about evaluating this integral.

Q: What is the significance of the limits of integration?

A: The limits of integration, $-1$ and $\sqrt{3}$, are crucial in evaluating the integral. The lower limit, $-1$, represents the starting point of the integral, while the upper limit, $\sqrt{3}$, represents the ending point. The choice of limits affects the value of the integral.

Q: Why is the integrand $\frac{1}{1+x^2}$ important?

A: The integrand $\frac{1}{1+x^2}$ is a rational function that can be expressed as the ratio of two polynomials. This function is important because it is a fundamental building block of many mathematical functions, including trigonometric functions and exponential functions.

Q: Can the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ be evaluated using other techniques?

A: Yes, the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ can be evaluated using other techniques, such as partial fractions or trigonometric substitution. However, the techniques used in this article, substitution and integration by parts, are two of the most common and effective methods.

Q: What is the relationship between the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ and the arctangent function?

A: The integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is related to the arctangent function, which is defined as $\arctan(x) = \int \frac{1}{1+x^2} \, dx$. This relationship is crucial in evaluating the integral.

Q: Can the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ be used to solve real-world problems?

A: Yes, the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ can be used to solve real-world problems, such as finding the area under curves or the volume of solids. This integral is a fundamental tool in many fields, including physics, engineering, and economics.

Q: What is the final answer to the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$?

A: The final answer to the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is $\boxed{\frac{7\pi}{12}}$.

Conclusion

In this article, we answered some common questions about evaluating the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$. We discussed the significance of the limits of integration, the importance of the integrand, and the relationship between the integral and the arctangent function. We also provided the final answer to the integral.

Final Answer

The final answer to the integral $\int_{-1}^{\sqrt{3}} \frac{1}{1+x^2} \, dx$ is $\boxed{\frac{7\pi}{12}}$.