Evaluate The Integral: ∫ ( X 2 + 2 − X ) D X \int\left(x^2+2^{-x}\right) D X ∫ ( X 2 + 2 − X ) D X

$$

Introduction

In this article, we will delve into the world of calculus and evaluate the given integral, $\int\left(x^2+2^{-x}\right) d x$. This problem requires us to apply various techniques from calculus, including substitution and integration by parts. Our goal is to find the antiderivative of the given function and express it in a simplified form.

Substitution Method

To evaluate the integral, we can start by using the substitution method. This method involves substituting a new variable into the integral to simplify it. In this case, we can let $u = 2^{-x}$, which implies that $du = -2^{-x} \ln 2 dx$. We can then rewrite the integral in terms of $u$.

Substitution

Let $u = 2^{-x}$. Then, $du = -2^{-x} \ln 2 dx$.

\int\left(x^2+2^{-x}\right) d x = \int\left(x^2 + u\right) \frac{1}{-u \ln 2} du

Integration by Parts

We can now use integration by parts to evaluate the integral. This method involves differentiating one function and integrating the other. In this case, we can let $v = x^2$ and $dw = \frac{1}{-u \ln 2} du$.

\int\left(x^2 + u\right) \frac{1}{-u \ln 2} du = \frac{x^2}{-u \ln 2} - \int \frac{2x}{-u \ln 2} du

Simplifying the Integral

We can simplify the integral by substituting back $u = 2^{-x}$.

\int\left(x^2 + u\right) \frac{1}{-u \ln 2} du = \frac{x^2}{-2^{-x} \ln 2} - \int \frac{2x}{-2^{-x} \ln 2} du

Evaluating the Integral

We can now evaluate the integral by simplifying the expression.

\int\left(x^2 + 2^{-x}\right) d x = \frac{x^2}{-2^{-x} \ln 2} + \frac{2x}{\ln 2} \ln 2^{-x} + C

Conclusion

In this article, we evaluated the given integral, $\int\left(x^2+2^{-x}\right) d x$, using the substitution method and integration by parts. We simplified the expression and found the antiderivative of the given function. The final answer is $\frac{x^2}{-2^{-x} \ln 2} + \frac{2x}{\ln 2} \ln 2^{-x} + C$.

Final Answer

The final answer is $\boxed{\frac{x^2}{-2^{-x} \ln 2} + \frac{2x}{\ln 2} \ln 2^{-x} + C}$.

Related Topics

• Integration by parts
• Substitution method
• Calculus

References

• [1] "Calculus" by Michael Spivak
• [2] "Calculus: Early Transcendentals" by James Stewart

Note: The references provided are for general calculus textbooks and are not specific to the problem at hand.

$$

Introduction

In our previous article, we evaluated the integral $\int\left(x^2+2^{-x}\right) d x$ using the substitution method and integration by parts. In this article, we will answer some common questions related to this problem.

Q1: What is the substitution method in calculus?

A1: The substitution method is a technique used in calculus to evaluate integrals by substituting a new variable into the integral. This method is useful when the integral contains a function that can be expressed in terms of the new variable.

Q2: How do I choose the substitution variable?

A2: When choosing a substitution variable, look for a function in the integral that can be expressed in terms of the new variable. For example, in the integral $\int\left(x^2+2^{-x}\right) d x$, we can let $u = 2^{-x}$ because $2^{-x}$ is a function of $x$.

Q3: What is integration by parts?

A3: Integration by parts is a technique used in calculus to evaluate integrals by differentiating one function and integrating the other. This method is useful when the integral contains a product of two functions.

Q4: How do I apply integration by parts?

A4: To apply integration by parts, let $v = x^2$ and $dw = \frac{1}{-u \ln 2} du$. Then, $dv = 2x dx$ and $w = -\frac{1}{u \ln 2}$. The integral becomes $\frac{x^2}{-u \ln 2} - \int \frac{2x}{-u \ln 2} du$.

Q5: What is the final answer to the integral $\int\left(x^2+2^{-x}\right) d x$?

A5: The final answer to the integral $\int\left(x^2+2^{-x}\right) d x$ is $\frac{x^2}{-2^{-x} \ln 2} + \frac{2x}{\ln 2} \ln 2^{-x} + C$.

Q6: What are some common mistakes to avoid when evaluating integrals?

A6: Some common mistakes to avoid when evaluating integrals include:

• Not choosing the correct substitution variable
• Not applying integration by parts correctly
• Not simplifying the expression after applying integration by parts
• Not including the constant of integration $C$

Q7: How do I practice evaluating integrals?

A7: To practice evaluating integrals, try solving problems from calculus textbooks or online resources. Start with simple problems and gradually move on to more complex ones. Make sure to check your work and simplify the expression after applying integration by parts.

Q8: What are some real-world applications of integrals?

A8: Integrals have many real-world applications, including:

• Physics: Integrals are used to calculate the area under curves, which represents the accumulation of a quantity over a given interval.
• Engineering: Integrals are used to calculate the volume of solids, which is essential in designing and building structures.
• Economics: Integrals are used to calculate the total cost or revenue of a business over a given period.

Conclusion

In this article, we answered some common questions related to evaluating the integral $\int\left(x^2+2^{-x}\right) d x$. We discussed the substitution method, integration by parts, and some common mistakes to avoid when evaluating integrals. We also provided some real-world applications of integrals and tips for practicing evaluating integrals.

Final Answer

The final answer is $\boxed{\frac{x^2}{-2^{-x} \ln 2} + \frac{2x}{\ln 2} \ln 2^{-x} + C}$.

Related Topics

• Integration by parts
• Substitution method
• Calculus

References

• [1] "Calculus" by Michael Spivak
• [2] "Calculus: Early Transcendentals" by James Stewart

Note: The references provided are for general calculus textbooks and are not specific to the problem at hand.