Question

Find: \((a) \int 13 e^{x} d x\) \((b) \int\left(3 e^{x}+\frac{4}{x}\right) d x \quad(x>0)\) \((c) \int\left(5 e^{x}+\frac{3}{x^{2}}\right) d x \quad(x \neq 0)\) \((d) \int 3 e^{-(2 x+7)} d x\) (e) \(\int 4 x e^{x^{2}+3} d x\) \((f) \int x e^{x^{2}}+9 d x\)

Short answer

(a) \( 13 e^{x} + C \), (b) \( 3 e^{x} + 4 \ln |x| + C \), (c) \( 5 e^{x} - \frac{3}{x} + C \), (d) \(-\frac{3}{2} e^{-(2x+7)} + C\), (e) \( e^{x^2+3} + C \), (f) \(\frac{1}{2} e^{x^2} + 9x + C\).

Step-by-step solution

Solve Part (a)

We need to solve \( \int 13 e^{x} \, dx \). The integral of \( e^x \) with respect to \( x \) is \( e^x \). Therefore, multiplying by the constant 13, we have:\[ \int 13 e^{x} \, dx = 13 e^{x} + C \]where \( C \) is the constant of integration.

Solve Part (b)

We need to solve \( \int \left(3 e^{x} + \frac{4}{x}\right) \, dx \). This can be split into two separate integrals:\[ \int 3 e^{x} \, dx + \int \frac{4}{x} \, dx \]The first integral \( \int 3 e^{x} \, dx = 3 e^{x} \). The second integral \( \int \frac{4}{x} \, dx = 4 \ln |x| \).Combining these results, we have:\[ \int \left(3 e^{x} + \frac{4}{x}\right) \, dx = 3 e^{x} + 4 \ln |x| + C \]

Solve Part (c)

We need to solve \( \int \left(5 e^{x} + \frac{3}{x^{2}}\right) \, dx \). This is also split into two integrals:\[ \int 5 e^{x} \, dx + \int \frac{3}{x^{2}} \, dx \]The first integral \( \int 5 e^{x} \, dx = 5 e^{x} \). The second integral can be rewritten as \( \int 3 x^{-2} \, dx \). The integral of \( x^{-n} \) is \( \frac{x^{-n+1}}{-n+1} \), so:\[ \int 3 x^{-2} \, dx = 3 \left( \frac{x^{-1}}{-1} \right) = -\frac{3}{x} \]Combining these results, we have:\[ \int \left(5 e^{x} + \frac{3}{x^{2}}\right) \, dx = 5 e^{x} - \frac{3}{x} + C \]

Solve Part (d)

We need to solve \( \int 3 e^{-(2x+7)} \, dx \). Using the substitution \( u = 2x + 7 \), \( du = 2 \, dx \), or \( dx = \frac{du}{2} \). The integral becomes:\[ \int 3 e^{-u} \cdot \frac{du}{2} = \frac{3}{2} \int e^{-u} \, du = -\frac{3}{2} e^{-u} + C \]Substituting back \( u = 2x + 7 \):\[ -\frac{3}{2} e^{-(2x+7)} + C \]

Solve Part (e)

We need to solve \( \int 4 x e^{x^{2}+3} \, dx \). Use substitution \( u = x^2 + 3 \), \( du = 2x \, dx \), or \( x \, dx = \frac{du}{2} \). The integral becomes:\[ 2 \int e^{u} \cdot \frac{du}{2} = \int e^{u} \, du = e^{u} + C \]Substituting back \( u = x^2 + 3 \):\[ e^{x^2 + 3} + C \]

Solve Part (f)

We need to solve \( \int (x e^{x^{2}} + 9) \, dx \). Split into two separate integrals:\[ \int x e^{x^{2}} \, dx + \int 9 \, dx \]For the first integral, use substitution \( u = x^2 \), \( du = 2x \, dx \). This becomes:\[ \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^{u} = \frac{1}{2} e^{x^2} \]The second integral \( \int 9 \, dx = 9x \).Combining these results, we have:\[ \frac{1}{2} e^{x^2} + 9x + C \]

Key concepts

Exponential Functions

Exponential functions are mathematical functions in which a constant base is raised to a variable exponent. These functions, often denoted as \( e^x \), where \( e \) is the base and \( x \) is the exponent, have unique properties that make them extensively used in calculus and other fields.

• Natural Exponential Function: The function \( e^x \) is known as the natural exponential function, where \( e \approx 2.71828 \). This function has the remarkable property that its derivative is the same as the function itself, making it straightforward to integrate.


• Application: Exponential functions are used to model growth and decay processes, such as population growth, radioactive decay, and compounded interest. Their ability to represent rapid changes makes them vital in many mathematical models.


• Integration: When integrating exponential functions like \( \int e^x \, dx \), the result is simply \( e^x + C \), where \( C \) is the constant of integration. If a constant multiplier is present, as seen in \( \int 13 e^x \, dx \), it can be factored out, simplifying the process to \( 13 \int e^x \, dx = 13e^x + C \).

The simplicity of working with exponential functions is due to their unique properties and wide applicability, making them an essential component of integral calculus.

Definite and Indefinite Integrals

Integration is a fundamental concept in calculus that helps us find areas under curves, among other things. Understanding definite and indefinite integrals is crucial in tackling various problems.

• Indefinite Integrals: These integrals, symbolized by \( \int f(x) \, dx \), represent a family of functions, including a constant of integration \( C \). They are used when we seek a general form of an antiderivative. For example, solving \( \int 3 e^x \, dx \) results in \( 3 e^x + C \), signifying that the integration represents a family of curves.


• Definite Integrals: These integrals, including limits \( \int_a^b f(x) \, dx \), offer numeric value representing the exact area under a curve between two limits \( a \) and \( b \). The main difference is that definite integrals do not include \( C \), given their numerical rendering.


• Key Properties: Properties like the linearity of integrals, \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \), allow flexibility in breaking problems into simpler parts for integration.

Understanding these concepts helps in solving an array of real-world problems by applying the correct integral form and identifying limits when needed.

Integration Techniques

Mastering integration often involves a variety of techniques to simplify or directly evaluate integrals, especially when faced with complicated expressions.

• Basic Integration: This approach involves simple direct integration, like with constant terms or basic functions, as seen in \( \int 13 e^{x} \, dx \). Recognizing direct antiderivatives is key.


• Substitution Method: When integrals involve more complex expressions, substitution can simplify them. For instance, in \( \int 4 x e^{x^2 + 3} \, dx \), setting \( u = x^2 + 3 \) allows us to convert it into a simpler integral of the form \( \int e^u \, du \).


• Breaking Down Sum/Difference: Splitting integrals into sums or differences, such as \( \int \left(3 e^x + \frac{4}{x} \right) \, dx \), allows us to independently integrate each term using known antiderivatives and generalized integration rules.

Employing these techniques systematically can streamline problem-solving in integral calculus, making it more approachable and less intimidating, even when the expressions seem complex at first glance.