Question

A useful integral a. Use integration by parts to show that if \(f^{\prime}\) is continuous, then $$\int x f^{\prime}(x) d x=x f(x)-\int f(x) d x$$. b. Use part (a) to evaluate \(\int x e^{3 x} d x\).

Short answer

Question: Prove the integral formula $\int x f'(x) dx = xf(x) - \int f(x) dx$ and use it to find $\int xe^{3x} dx$. Solution: Step 1: Recall the integration by parts formula $$ \int u dv = u v - \int v d u $$ Step 2: Define u and dv for our integral According to the given integral, we have: $$ \int x f^{\prime}(x) dx$$ We can choose u and dv as: $$u = x,\ du = \frac{d(x)}{dx}dx = dx$$ and $$dv = f^{\prime}(x)dx,\ v =\int f^{\prime}(x)dx =f(x)$$ Step 3: Apply the integration by parts Applying the integration by parts for u and dv defined in step 2, we get: $$\int x f^{\prime}(x) d x = x f(x) - \int f(x) d x$$ Thus, we have proved the given formula in part (a). Step 4: Use part (a) results to solve the given integral Let's use the result from part (a) to evaluate the integral: $$\int x e^{3 x} d x$$ Here, we have \(f(x) = e^{3x}\), so \(f^{\prime}(x) = 3e^{3x}\). Step 5: Substitute f(x) and f'(x) in the integral formula Substitute \(f(x)\) and \(f^{\prime}(x)\) in the formula from part (a), we get: $$\int x f^{\prime}(x) d x = xf(x) - \int f(x) d x$$ $$\int x (3e^{3x}) d x = x(e^{3x}) - \int e^{3x} d x$$ Step 6: Integrate the remaining part Now, we need to find the integral of \(e^{3x}\): $$\int e^{3x} d x = \frac{e^{3x}}{3}$$ Step 7: Write the final solution Finally, we substitute the integral of \(e^{3x}\) into the formula: $$\int x e^{3 x} d x = xe^{3 x} - \int e^{3 x} d x = xe^{3x} - \frac{e^{3x}}{3} + C$$ Hence, the integral evaluates to: $$\int x e^{3 x} d x = xe^{3x} - \frac{e^{3x}}{3} + C$$ where C is the constant of integration.

Step-by-step solution

Recall the integration by parts formula

The integration by parts formula can be written as: $$ \int u dv = u v - \int v d u $$

Define u and dv for our integral

According to the given integral, we have: $$ \int x f^{\prime}(x) dx$$ So, we can choose u and dv as: $$u = x,\ du = \frac{d(x)}{dx}dx = dx$$ and $$dv = f^{\prime}(x)dx,\ v =\int f^{\prime}(x)dx =f(x)$$

Apply the integration by parts

Applying the integration by parts for u and dv defined in step 2, we get: $$\int x f^{\prime}(x) d x = x f(x) - \int f(x) d x$$ Thus, we have proved the given formula in part (a). #Part b#

Use part (a) results to solve the given integral

Let's use the result from part (a) to evaluate the integral: $$\int x e^{3 x} d x$$ Here, we have \(f(x) = e^{3x}\), so \(f^{\prime}(x) = 3e^{3x}\).

Substitute f(x) and f'(x) in the integral formula

Substitute \(f(x)\) and \(f^{\prime}(x)\) in the formula from part (a), we get: $$\int x f^{\prime}(x) d x = xf(x) - \int f(x) d x$$ $$\int x (3e^{3x}) d x = x(e^{3x}) - \int e^{3x} d x$$

Integrate the remaining part

Now, we need to find the integral of \(e^{3x}\): $$\int e^{3x} d x = \frac{e^{3x}}{3}$$

Write the final solution

Finally, we substitute the integral of \(e^{3x}\) into the formula: $$\int x e^{3 x} d x = xe^{3 x} - \int e^{3 x} d x = xe^{3x} - \frac{e^{3x}}{3} + C$$ Hence, the integral evaluates to: $$\int x e^{3 x} d x = xe^{3x} - \frac{e^{3x}}{3} + C$$ where C is the constant of integration.

Key concepts

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is fundamental in calculus. An indefinite integral is essentially the antiderivative of a function and is written without limits of integration. It represents a family of functions and includes a constant of integration, denoted as 'C'. For instance, if we integrate the function \( f(x) \), the result is the general form \( F(x) + C \), where \( F(x) \) is any antiderivative of \( f(x) \).

Definite integrals, on the other hand, are written with limits of integration and give a numerical result. They calculate the net area under the curve of a function between two points on the x-axis, labeled as the lower \( a \) and upper \( b \) limits: \( \int_{a}^{b} f(x) dx \). The definite integral evaluates to \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \).

When solving problems like the exercise above using integration by parts, we're often dealing with indefinite integrals to find a general formula that we can later apply to specific cases, such as when evaluating a definite integral.

Integrating Exponential Functions

Working with exponential functions is a common scenario in integration. Exponential functions are of the form \( e^{kx} \), where \( k \) is a constant. When integrating an exponential function, the antiderivative is remarkably similar to the original function: \( \int e^{kx} dx = \frac{1}{k}e^{kx} + C \), where 'C' represents the constant of integration.

In our exercise, we integrated \( x e^{3x} \) by applying integration by parts. Here, we consider \( e^{3x} \) as part of the product to be integrated, leading to the simple antiderivative \( \frac{1}{3}e^{3x} \). Integrating exponential functions is straightforward because their rate of change is proportional to their value, making them indispensable in modeling growth and decay processes in various scientific fields.

Continuous Functions

The concept of continuity is essential when discussing the integrability of functions. A function is said to be continuous if it has no breaks, holes, or jumps. Mathematically, a function \( f(x) \) is continuous at a point \( x = c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). For the function to be continuous on an interval, it must be continuous at every point within that interval.

Continuous functions have the important property that they can be integrated over an interval. The exercise assumes that \( f'(x) \) is continuous, which guarantees the validity of the integration by parts method. This requirement is fundamental to ensure the existence of a well-defined antiderivative that can be applied to the integration process.

Product Rule for Derivatives

Integration by parts is closely related to the product rule for derivatives, which tells us how to find the derivative of the product of two functions. The product rule states that if we have two differentiable functions \( u(x) \) and \( v(x) \), then the derivative of their product is \( u'v + uv' \).

From the product rule, by algebraic manipulation and integrating both sides, we arrive at the integration by parts formula: \( \int u dv = uv - \int v du \). It is a fundamental technique used to integrate products of functions and is particularly useful when one function is the derivative of another. In our exercise, \( x \) and \( f^{\textprime}(x) \) are chosen to be \( u \) and \( dv \) respectively, and applying the integration by parts technique reduces the original integral to a simpler form that can be easily integrated.