Problem 63
Question
Right, or wrong? Say which for each formula and give a brief reason for each answer. a. \(\int x \sin x d x=\frac{x^{2}}{2} \sin x+C\) b. \(\int x \sin x d x=-x \cos x+C\) c. \(\int x \sin x d x=-x \cos x+\sin x+C\)
Step-by-Step Solution
Verified Answer
a. Wrong (derivative doesn't match).
b. Wrong (derivative doesn't match).
c. Correct (derivative matches original).
1Step 1: Understanding the Problem
We are tasked with evaluating whether the given antiderivative formulas for the integral \( \int x \sin x \, dx \) are correct and providing a brief reason for each. To solve this, we will check each provided formula by differentiating the proposed antiderivative and comparing it to the original integrand \( x \sin x \).
2Step 2: Evaluate Formula (a)
The given formula is \( \int x \sin x \, dx = \frac{x^{2}}{2} \sin x + C \). Differentiate the right-hand side: \[ \frac{d}{dx} \left( \frac{x^{2}}{2} \sin x + C \right) = \frac{x^{2}}{2} \cos x + x \sin x \]The derivative is not equal to the original integrand \( x \sin x \). Thus, formula (a) is wrong.
3Step 3: Evaluate Formula (b)
The given formula is \( \int x \sin x \, dx = -x \cos x + C \). Differentiate the right-hand side: \[ \frac{d}{dx} (-x \cos x + C) = -\cos x + x \sin x \] The derivative \( -\cos x + x \sin x \) differs from \( x \sin x \), so formula (b) is wrong.
4Step 4: Evaluate Formula (c)
The given formula is \( \int x \sin x \, dx = -x \cos x + \sin x + C \). Differentiate the right-hand side: \[ \frac{d}{dx} \left( -x \cos x + \sin x + C \right) = -\cos x + x \sin x + \cos x = x \sin x \] The derivative simplifies to \( x \sin x \), matching the original integrand. Thus, formula (c) is correct.
Key Concepts
AntiderivativesDifferentiationIntegral Calculus
Antiderivatives
Antiderivatives, also known as indefinite integrals, are an essential concept in calculus. An antiderivative is a function whose derivative is equal to the given function. In simpler terms, it is the reverse process of differentiation. Finding an antiderivative allows us to determine the original function based on its rate of change.
For a function \( f(x) \), the antiderivative is typically expressed as \( F(x) + C \), where \( C \) is the constant of integration. This constant accounts for the fact that there are infinitely many antiderivatives that differ by a constant value. Hence, to check whether a given integral solution is correct, one often differentiates it to see if it returns the original function.
By recognizing incorrect antiderivatives, as in formulas (a) and (b) from the exercise, we ensure that our understanding of antiderivatives and the integral calculus is precise.
For a function \( f(x) \), the antiderivative is typically expressed as \( F(x) + C \), where \( C \) is the constant of integration. This constant accounts for the fact that there are infinitely many antiderivatives that differ by a constant value. Hence, to check whether a given integral solution is correct, one often differentiates it to see if it returns the original function.
By recognizing incorrect antiderivatives, as in formulas (a) and (b) from the exercise, we ensure that our understanding of antiderivatives and the integral calculus is precise.
Differentiation
Differentiation is the process of finding the derivative of a function, which measures how the function value changes as its input changes. In the context of antiderivatives, differentiation is used to verify if the proposed solution to an integral is correct.
By differentiating potential solutions, we can see whether the solution yields the original function before integration. This checking process is essential to confirm the accuracy of integration by parts and other techniques used to solve integrals.
By differentiating potential solutions, we can see whether the solution yields the original function before integration. This checking process is essential to confirm the accuracy of integration by parts and other techniques used to solve integrals.
- It involves applying rules such as the power rule, product rule, and chain rule to the function.
- In our exercise, each proposed solution was differentiated and compared with \( x \sin x \).
- Only the correct differentiation process will bring the original integrand, affirming the accuracy of the proposed solution.
Integral Calculus
Integral calculus is a vast area of mathematics that focuses on the concept of integration. It involves finding the total accumulation of quantities, like areas under curves, given a rate of change, such as velocity or force.
In the exercise, we are dealing with a specific integration technique called "Integration by Parts." This method is especially useful when the integrand is a product of two functions, like \( x \sin x \).
In the exercise, we are dealing with a specific integration technique called "Integration by Parts." This method is especially useful when the integrand is a product of two functions, like \( x \sin x \).
- Integration by parts is based on the product rule for differentiation and uses the formula:
\[ \int u \, dv = uv - \int v \, du \]. - Selecting appropriate \( u \) and \( dv \) is crucial for simplifying the integration process.
- This technique allows us to transform the integral into a more manageable form.
Other exercises in this chapter
Problem 62
Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2} \sqrt{3-x}$$
View solution Problem 62
Show that the functions in Exercises 61 and 62 have local extreme values at the given values of \(\theta\), and say which kind of local extreme the function has
View solution Problem 63
Show that \(|\cos x-1| \leq|x|\) for all \(x\) -values.
View solution Problem 63
Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=\left\\{\begin{array}{ll} 4-2 x, & x \leq 1 \\ x+1, &
View solution