What is the anti-derivative of e to the kx?

The integral of e to the power kx with respect to x is one over k times e to the power kx, plus the constant of integration C. You divide by the coefficient of x because differentiating e to the kx multiplies by k, so reversing it divides by k. For example the integral of e to the 5x is one fifth e to the 5x plus C.

How do you integrate e to the power ax plus b?

Only the coefficient of x affects the factor: the integral of e to the power ax plus b is one over a times e to the power ax plus b, plus C. The constant b inside the exponent is left unchanged. For example the integral of e to the 4x minus 3 is one quarter e to the 4x minus 3, plus C.

How do you evaluate a definite integral of e to the kx?

Find the anti-derivative F of x equals one over k times e to the kx, then apply the fundamental theorem of calculus: the integral from a to b equals F of b minus F of a. For example the integral from 0 to 1 of e to the 2x is one half e squared minus one half, which is one half times e squared minus 1.

How do you find the constant of integration from a boundary point?

Anti-differentiate to get the function with an unknown constant C, then substitute the given point. If f dashed of x equals e to the 2x and f of 0 equals 3, then f of x equals one half e to the 2x plus C, and setting x equals 0 gives one half plus C equals 3, so C equals five halves.

What is the anti-derivative of e to the kx?

The integral of e to the power kx with respect to x is one over k times e to the power kx, plus the constant of integration C. You divide by the coefficient of x because differentiating e to the kx multiplies by k, so reversing it divides by k. For example the integral of e to the 5x is one fifth e to the 5x plus C.

How do you integrate e to the power ax plus b?

Only the coefficient of x affects the factor: the integral of e to the power ax plus b is one over a times e to the power ax plus b, plus C. The constant b inside the exponent is left unchanged. For example the integral of e to the 4x minus 3 is one quarter e to the 4x minus 3, plus C.

How do you evaluate a definite integral of e to the kx?

Find the anti-derivative F of x equals one over k times e to the kx, then apply the fundamental theorem of calculus: the integral from a to b equals F of b minus F of a. For example the integral from 0 to 1 of e to the 2x is one half e squared minus one half, which is one half times e squared minus 1.

How do you find the constant of integration from a boundary point?

Anti-differentiate to get the function with an unknown constant C, then substitute the given point. If f dashed of x equals e to the 2x and f of 0 equals 3, then f of x equals one half e to the 2x plus C, and setting x equals 0 gives one half plus C equals 3, so C equals five halves.

Year 12 Methods (Unit 3 & 4) Anti-differentiation

The anti-derivative of e^kx

Q: Why do you divide by k instead of multiplying?

Because differentiation and anti-differentiation are opposite operations. Differentiating e to the kx multiplies by k by the chain rule, so anti-differentiating must divide by k to undo it. Multiplying by k instead would differentiate to k squared times e to the kx, which is wrong.

Q: Do you always need the constant of integration?

Yes, for an indefinite integral. Because differentiating any constant gives zero, the anti-derivative is a whole family of curves differing by a constant, so you must write plus C. Only a definite integral, or a boundary condition that fixes C, removes the arbitrary constant.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), anti-differentiation reverses differentiation. Because \(\dfrac{d}{dx}e^{kx}=k\,e^{kx}\), the anti-derivative of \(e^{kx}\) divides by \(k\): \(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\). This page covers the base rule, the linear-inner form \(e^{ax+b}\), linearity, definite integrals by the fundamental theorem, and the particular anti-derivative from a boundary point.

An anti-derivative of a function \(f\) is a function \(F\) whose derivative is \(f\); anti-differentiation is the reverse of differentiation. For the exponential function base \(e\), differentiation multiplies by the coefficient of \(x\), so anti-differentiation divides by it.

The core result is

\(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C.\)

More generally, for a linear inner function \(ax+b\), only the coefficient of \(x\) sets the factor: \(\displaystyle\int e^{ax+b}\,dx=\dfrac{1}{a}e^{ax+b}+C\). Using linearity you integrate a sum term by term, and a definite integral is found by evaluating the anti-derivative at the two limits. Because differentiating any constant gives \(0\), an indefinite integral is a whole family of curves, and a boundary point \(f(a)=b\) fixes the constant \(C\) to pick out one particular anti-derivative.

Key idea. \(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\): divide by \(k\), never multiply, and always add \(+C\). For \(e^{ax+b}\), divide by \(a\) and leave \(b\) unchanged.

Definite integral: the shaded area is \(\displaystyle\int_{0}^{1} e^{x}\,dx=e-1\approx 1.72\)

Boundary point: \(f'(x)=e^{2x}\), \(f(0)=3\) fixes \(C\), giving \(f(x)=\dfrac{1}{2}e^{2x}+\dfrac{5}{2}\)

The base rule (anti-derivative of \(e^{kx}\)):

\[\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\]

\int e^{k x} d x = \frac{1}{k} e^{k x} + C

The linear-inner form and linearity:

\[\int e^{ax+b}\,dx=\dfrac{1}{a}e^{ax+b}+C,\qquad \int\!\big(f+g\big)\,dx=\int f\,dx+\int g\,dx,\qquad \int k\,f\,dx=k\int f\,dx\]

\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + C

A definite integral, by the fundamental theorem of calculus:

\[\int_{a}^{b} e^{kx}\,dx=\Big[\dfrac{1}{k}e^{kx}\Big]_{a}^{b}=F(b)-F(a)\]

\int_{a}^{b} e^{k x} d x = F (b) - F (a)

Boundary point. Given \(f'(x)\) and a condition \(f(a)=b\), anti-differentiate to get \(f(x)=\ldots+C\), then substitute the point to solve for \(C\) — this selects the one particular anti-derivative.

How to anti-differentiate an exponential

Identify the coefficient of \(x\). For \(e^{kx}\) it is \(k\); for \(e^{ax+b}\) it is \(a\) (ignore the constant \(b\)).
Divide by that coefficient. Write \(\dfrac{1}{k}e^{kx}\) — divide, never multiply, and watch the sign when the coefficient is negative.
Add \(+C\). Every indefinite integral needs the constant of integration.
Use linearity for sums. Integrate each term separately and combine, with a single \(C\).
Definite or particular. For a definite integral evaluate \(F(b)-F(a)\); for a particular anti-derivative substitute the boundary point to find \(C\).

Check by differentiating. A quick way to be sure: differentiate your answer. \(\dfrac{d}{dx}\!\left(\dfrac{1}{k}e^{kx}\right)=\dfrac{1}{k}\cdot k\,e^{kx}=e^{kx}\), which returns the original integrand.

Example 1 — Base rule

Find \(\displaystyle\int e^{5x}\,dx\).

Solution

Divide by the coefficient \(k=5\) and add \(C\).

\(\displaystyle\int e^{5x}\,dx\)

\(=\)

\(\dfrac{1}{5}e^{5x}+C\)

\int e^{5 x} d x = \frac{1}{5} e^{5 x} + C

Example 2 — Linear inner

Find \(\displaystyle\int e^{4x-3}\,dx\).

Solution

Only the coefficient of \(x\) matters: divide by \(a=4\), leave \(-3\) unchanged.

\(\displaystyle\int e^{4x-3}\,dx\)

\(=\)

\(\dfrac{1}{4}e^{4x-3}+C\)

\int e^{4 x - 3} d x = \frac{1}{4} e^{4 x - 3} + C

Example 3 — Definite integral

Evaluate \(\displaystyle\int_{0}^{1} e^{2x}\,dx\).

Solution

Anti-differentiate to \(\dfrac{1}{2}e^{2x}\), then apply \(F(1)-F(0)\).

\(\displaystyle\int_{0}^{1} e^{2x}\,dx\)	\(=\)	\(\Big[\dfrac{1}{2}e^{2x}\Big]_{0}^{1}\)
	\(=\)	\(\dfrac{1}{2}e^{2}-\dfrac{1}{2}\)
	\(=\)	\(\dfrac{1}{2}\big(e^{2}-1\big)\)

\int_{0}^{1} e^{2 x} d x = \frac{1}{2} (e^{2} - 1)

Example 4 — Boundary point

\(f'(x)=e^{2x}\) and \(f(0)=3\). Find \(f(x)\).

Solution

Anti-differentiate, then use \(f(0)=3\) (with \(e^{0}=1\)) to find \(C\).

\(f(x)\)	\(=\)	\(\dfrac{1}{2}e^{2x}+C\)
\(\dfrac{1}{2}+C\)	\(=\)	\(3\)
\(C\)	\(=\)	\(\dfrac{5}{2}\)
\(f(x)\)	\(=\)	\(\dfrac{1}{2}e^{2x}+\dfrac{5}{2}\)

f (x) = \frac{1}{2} e^{2 x} + \frac{5}{2}

Common pitfalls

Divide by \(k\), do not multiply. \(\displaystyle\int e^{5x}\,dx=\dfrac{1}{5}e^{5x}+C\), not \(5e^{5x}+C\). Multiplying reverses the operation the wrong way.

Do not drop the \(\dfrac{1}{k}\). Writing \(\displaystyle\int e^{5x}\,dx=e^{5x}+C\) is wrong — that differentiates back to \(5e^{5x}\), not \(e^{5x}\).

Mind the sign when \(k<0\). \(\displaystyle\int e^{-x}\,dx=\dfrac{1}{-1}e^{-x}+C=-e^{-x}+C\).

Never forget \(+C\). An indefinite integral is a family of curves; leaving off the constant loses every anti-derivative but one.

Frequently asked questions

What is the anti-derivative of e^kx?

\(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\). You divide by \(k\) because differentiating \(e^{kx}\) multiplies by \(k\); e.g. \(\displaystyle\int e^{5x}\,dx=\dfrac{1}{5}e^{5x}+C\).

How do you integrate e^(ax+b)?

Only the coefficient of \(x\) sets the factor: \(\displaystyle\int e^{ax+b}\,dx=\dfrac{1}{a}e^{ax+b}+C\); the constant \(b\) is unchanged. E.g. \(\displaystyle\int e^{4x-3}\,dx=\dfrac{1}{4}e^{4x-3}+C\).

Why do you divide by k instead of multiplying?

Anti-differentiation undoes differentiation. The chain rule multiplies \(e^{kx}\) by \(k\), so reversing it divides by \(k\). Multiplying would differentiate to \(k^{2}e^{kx}\).

How do you evaluate a definite integral of e^kx?

Use \(F(b)-F(a)\) with \(F(x)=\dfrac{1}{k}e^{kx}\). E.g. \(\displaystyle\int_{0}^{1} e^{2x}\,dx=\dfrac{1}{2}e^{2}-\dfrac{1}{2}=\dfrac{1}{2}(e^{2}-1)\).

How do you find C from a boundary point?

Anti-differentiate to \(f(x)=\ldots+C\), then substitute the given point. For \(f'(x)=e^{2x}\), \(f(0)=3\): \(\dfrac{1}{2}+C=3\), so \(C=\dfrac{5}{2}\).

Do you always need the constant of integration?

Yes for an indefinite integral, because any constant differentiates to \(0\). A definite integral or a boundary condition removes the arbitrary \(C\).

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