Integration of Exponential Functions Revision Notes for HSC SSCE Mathematics Advanced

Integration of Exponential Functions

Introduction

Integration is the reverse process of differentiation. When we find a primitive (or antiderivative), we're working backwards from a derivative to find the original function. Since we know how to differentiate exponential functions, we can reverse these rules to create standard forms for integrating exponential functions.

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The Key Relationship

Integration and differentiation are inverse operations. This means that any differentiation rule can be reversed to create an integration formula. Understanding differentiation of exponentials is therefore essential for mastering their integration.

Standard forms for integration

By reversing the differentiation formulas for exponential functions, we obtain two essential integration formulas.

Basic exponential integral

Since $\frac{d}{dx}e^x = e^x$ , reversing this process gives us:

$\int e^x \, dx = e^x + C$

where $C$ is an arbitrary constant of integration.

Linear exponential integral

Since $\frac{d}{dx}e^{ax+b} = ae^{ax+b}$ , reversing this gives us:

$\int ae^{ax+b} \, dx = e^{ax+b}$

Dividing both sides by $a$ , we get:

$\int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + C$

where $a$ and $b$ are constants, and $C$ is the constant of integration.

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Dividing by the coefficient

Notice that the coefficient of $x$ in the exponent becomes the denominator in the result. This is a crucial pattern to remember: when you see $e^{ax+b}$ , you must divide by $a$ when integrating.

Reverse chain rule (reference)

Although not required in the SSCE course, there's also a reverse chain rule formula for exponential functions:

$\int e^u \frac{du}{dx} \, dx = e^u + C$

or equivalently:

$\int e^{f(x)} f'(x) \, dx = e^{f(x)} + C$

This can be useful for more complex integration problems.

Working with indefinite integrals

Indefinite integrals represent the general antiderivative of a function, always including the constant of integration $C$ .

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Worked Example: Finding indefinite integrals

Let's find these indefinite integrals:

Part a: $\int e^{3x+2} \, dx$

Using the standard form with $a = 3$ and $b = 2$ :

$\int e^{3x+2} \, dx = \frac{1}{3}e^{3x+2} + C$

Part b: $\int (1 - x + e^x) \, dx$

We integrate each term separately:

$\int (1 - x + e^x) \, dx = x - \frac{1}{2}x^2 + e^x + C$

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Integrating sums and differences

When integrating a sum or difference of terms, you can integrate each term separately and combine the results. This is a direct consequence of the linearity of integration.

Evaluating definite integrals

Definite integrals have upper and lower limits. We evaluate them by finding the primitive function, then substituting the limits and calculating the difference.

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Worked Example: Evaluating definite integrals

Part a: Evaluate $\int_0^2 e^x \, dx$

First, find the primitive: $\int e^x \, dx = e^x$

Now substitute the limits:

$\int_0^2 e^x \, dx = \left[e^x\right]_0^2$

$= e^2 - e^0$

$= e^2 - 1$

Part b: Evaluate $\int_2^3 e^{5-2x} \, dx$

Using the standard form with $a = -2$ and $b = 5$ :

$\int_2^3 e^{5-2x} \, dx = -\frac{1}{2}\left[e^{5-2x}\right]_2^3$

$= -\frac{1}{2}(e^{-1} - e^1)$

$= -\frac{1}{2}\left(\frac{1}{e} - e\right)$

$= \frac{e^2 - 1}{2e}$

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Watch out for negative coefficients!

When the coefficient of $x$ is negative, don't forget the negative sign in your working. For example, with $e^{5-2x}$ , you must use $a = -2$ , giving $\frac{1}{-2} = -\frac{1}{2}$ .

Finding functions from derivatives

When we know the derivative of a function and the value of the function at one specific point, we can determine the complete original function. This process involves finding the constant of integration using the given boundary condition.

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Worked Example: Finding the original function

Given that $f'(x) = e^x$ and $f(1) = 0$ :

Part a: Find the original function $f(x)$

We're given that $f'(x) = e^x$

Taking the primitive (integrating):

$f(x) = e^x + C$

where $C$ is some constant.

To find $C$ , we use the condition $f(1) = 0$ :

Substituting $x = 1$ :

$0 = e^1 + C$

$C = -e$

Therefore:

$f(x) = e^x - e$

Part b: Hence find $f(0)$

Substituting $x = 0$ into our function:

$f(0) = e^0 - e$

$= 1 - e$

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Worked Example: Working with more complex derivatives

If $f'(x) = 1 + 2e^{-x}$ and $f(0) = 1$ , find $f(x)$ and $f(1)$ .

Part a: Find $f(x)$

We're given that $f'(x) = 1 + 2e^{-x}$

Taking the primitive:

$f(x) = x - 2e^{-x} + C$

Note: For $2e^{-x}$ , we use the standard form with $a = -1$ , giving $\frac{2}{-1}e^{-x} = -2e^{-x}$

Using the condition $f(0) = 1$ :

$1 = 0 - 2e^0 + C$

$1 = 0 - 2 + C$

$C = 3$

Therefore:

$f(x) = x - 2e^{-x} + 3$

Part b: Find $f(1)$

Substituting $x = 1$ :

$f(1) = 1 - 2e^{-1} + 3$

$= 4 - 2e^{-1}$

Finding integrals from derivatives

Any result from differentiation can be reversed to give us an integration result. This technique is particularly useful for finding new primitives and demonstrates the powerful relationship between these two operations.

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Worked Example: Using differentiation to find an integral

Part a: Use the chain rule to differentiate $e^{x^2}$

Let $y = e^{x^2}$ and $u = x^2$

Then $y = e^u$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

We have:

$\frac{du}{dx} = 2x$

$\frac{dy}{du} = e^u$

Therefore:

$\frac{dy}{dx} = 2xe^{x^2}$

Part b: Hence find $\int_{-1}^1 2xe^{x^2} \, dx$

From part a, we know that $\frac{d}{dx}e^{x^2} = 2xe^{x^2}$

Reversing this to give a primitive:

$\int 2xe^{x^2} \, dx = e^{x^2}$

Therefore:

$\int_{-1}^1 2xe^{x^2} \, dx = \left[e^{x^2}\right]_{-1}^1$

$= e^1 - e^1$

$= :success[0]$

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Understanding odd functions

This result could have been predicted without calculation. The function $f(x) = xe^{x^2}$ is an odd function because:

$f(-x) = (-x)e^{(-x)^2} = -xe^{x^2} = -f(x)$

When an odd function is integrated over a symmetric interval $[-a, a]$ , the result is always zero. This happens because the positive and negative areas cancel out perfectly.

Using the reverse chain rule

For more complex exponential integrals, we can use the reverse chain rule formula to handle compositions of functions.

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Worked Example: Applying the reverse chain rule

Find $\int \frac{e^{2x}}{(1-e^{2x})^3} \, dx$

We can rewrite this as:

$\int \frac{e^{2x}}{(1-e^{2x})^3} \, dx = -\frac{1}{2} \int \frac{-2e^{2x}}{(1-e^{2x})^3} \, dx$

Let $u = 1 - e^{2x}$ , so $u' = -2e^{2x}$

Alternatively, let $f(x) = 1 - e^{2x}$ , so $f'(x) = -2e^{2x}$

Using the reverse chain rule:

$\int u^{-3} \frac{du}{dx} \, dx = \frac{u^{-2}}{-2}$

or equivalently:

$\int (f(x))^{-3} f'(x) \, dx = \frac{(f(x))^{-2}}{-2}$

Therefore:

$= -\frac{1}{2} \times \left(-\frac{1}{2}\right) \times (1-e^{2x})^{-2}$

$= \frac{1}{4(1-e^{2x})^2} + C$

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Checking for the right form

When using the reverse chain rule, always check that your integrand contains both a function and its derivative (or a constant multiple of its derivative). If this pattern isn't present, the reverse chain rule cannot be applied directly.

Remember!

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Key Points to Remember:

The two standard forms are: $\int e^x \, dx = e^x + C$ and $\int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + C$
Always divide by the coefficient of $x$ when integrating $e^{ax+b}$
Never forget the constant of integration $C$ for indefinite integrals (but don't include it for definite integrals)
When evaluating definite integrals, always subtract the lower limit value from the upper limit value: $[F(x)]_a^b = F(b) - F(a)$
The reverse chain rule can help with complex exponentials, but check that your integrand has the right form first

Integration of Exponential Functions (HSC SSCE Mathematics Advanced): Revision Notes