A biologist is tracking tumour growth. The growth rate is proportional to the tumour's current volume: $\frac{dV}{dt} = 0.1V$. To find the total volume over time, she needs to integrate an exponential function. A climate scientist measures atmospheric CO₂ absorption, which follows a logarithmic pattern. These two functions — exponential and logarithmic — are the natural language of growth and accumulation. This lesson teaches you to integrate them both.
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We know $\frac{d}{dx}(e^x) = e^x$. What do you think $\int e^x \, dx$ equals? And what about $\int \frac{1}{x} \, dx$?
Core Content
Since $\frac{d}{dx}(e^x) = e^x$, it follows immediately that:
∫e^x dx = e^x + C
For $e^{kx}$, we use the chain rule in reverse:
∫e^{kx} dx = (1/k)e^{kx} + C
Check: $\frac{d}{dx}(\frac{1}{k}e^{kx}) = \frac{1}{k} \cdot k \cdot e^{kx} = e^{kx}$ ✓
Examples:
The power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ fails for $n = -1$ because the denominator becomes zero.
But we know $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$. For $x < 0$, $\frac{d}{dx}(\ln(-x)) = \frac{-1}{-x} = \frac{1}{x}$.
Combining both cases:
∫(1/x) dx = ln|x| + C
The absolute value is essential. $\ln x$ is only defined for $x > 0$, but $\frac{1}{x}$ exists for all $x \neq 0$. The absolute value lets us cover both positive and negative domains.
Find $\int (2e^{3x} + \frac{4}{x}) \, dx$.
The antiderivative.
∫2e^{3x} dx = 2 · (1/3)e^{3x} = (2/3)e^{3x}
∫(4/x) dx = 4ln|x|
Combine: (2/3)e^{3x} + 4ln|x| + C
$\frac{2}{3}e^{3x} + 4\ln|x| + C$.
Find $\int (e^{-x} + \frac{3}{x} + x^2) \, dx$.
Answer:
$-e^{-x} + 3\ln|x| + \frac{x^3}{3} + C$.
Since $\frac{d}{dx}(a^x) = a^x \ln a$, dividing by $\ln a$ gives:
∫a^x dx = a^x / ln(a) + C
Check: $\frac{d}{dx}(\frac{a^x}{\ln a}) = \frac{a^x \ln a}{\ln a} = a^x$ ✓
Example: $\int 2^x \, dx = \frac{2^x}{\ln 2} + C$.
This is less common than $e^x$ because any $a^x$ can be rewritten as $e^{x \ln a}$.
$\int e^x \, dx = e^x + C$
$\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C$
$\int a^x \, dx = \frac{a^x}{\ln a} + C$
$\int \frac{1}{x} \, dx = \ln|x| + C$
Differentiate your answer
$\int e^x \, dx = e^x + C$ — the function that equals its own derivative also equals its own integral. $\int \frac{1}{x} \, dx = \ln|x| + C$ — this is the special case that the power rule cannot handle, and it is why logarithms appear naturally in integration. The absolute value ensures the antiderivative is valid for both positive and negative $x$.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find $\int (3e^{2x} - \frac{5}{x} + 4x) \, dx$. Show all working. 3 MARKS
9. Find $\int (e^{-x} + x^{-1} + x^{1/2}) \, dx$. Show all working. 3 MARKS
10. A radioactive tracer has decay rate $\frac{dA}{dt} = -2e^{-0.5t}$ mg/h. Find $A(t)$ given $A(0) = 4$. Calculate the amount remaining after 4 hours and explain what happens to $A(t)$ as $t \to \infty$. Sketch the graph of $A(t)$ showing the initial amount, the amount at $t = 4$, and the long-term behaviour. 3 MARKS
1. $\frac{1}{4}e^{4x} + C$
2. $-\frac{5}{2}e^{-2x} + C$
3. $3\ln|x| + C$
4. $e^x + \ln|x| + \frac{x^4}{4} + C$
1. The power rule gives division by zero. Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$, the integral is $\ln|x| + C$.
2. $\frac{d}{dx}(\frac{3^x}{\ln 3}) = \frac{3^x \ln 3}{\ln 3} = 3^x$.
3. $A = \int -0.5e^{-0.5t} \, dt = e^{-0.5t} + C$. At $t = 0$: $2 = 1 + C$, so $C = 1$. $A(t) = e^{-0.5t} + 1$.
Q8 (3 marks): $\int 3e^{2x} \, dx = \frac{3}{2}e^{2x}$ [0.5], $\int (-\frac{5}{x}) \, dx = -5\ln|x|$ [0.5], $\int 4x \, dx = 2x^2$ [0.5]. Answer: $\frac{3}{2}e^{2x} - 5\ln|x| + 2x^2 + C$ [1.5].
Q9 (3 marks): $\int e^{-x} \, dx = -e^{-x}$ [0.5], $\int x^{-1} \, dx = \ln|x|$ [0.5], $\int x^{1/2} \, dx = \frac{2}{3}x^{3/2}$ [0.5]. Answer: $-e^{-x} + \ln|x| + \frac{2}{3}x^{3/2} + C$ [1.5].
Q10 (3 marks): $A(t) = \int -2e^{-0.5t} \, dt = 4e^{-0.5t} + C$ [0.5]. $A(0) = 4$ gives $4 = 4 + C$, so $C = 0$ [0.25]. $A(t) = 4e^{-0.5t}$ [0.25]. At $t = 4$: $A(4) = 4e^{-2} = 4/e^2 \approx 0.541$ mg [0.5]. As $t \to \infty$, $e^{-0.5t} \to 0$, so $A(t) \to 0$ [0.5]. The tracer is completely eliminated from the body in the long run [0.25]. Sketch: starts at $(0, 4)$, decreases exponentially, passes through $(4, 0.54)$, asymptotically approaches $A = 0$ [0.75].
Climb platforms using e^x, 1/x, and mixed integration. Pool: lesson 3.
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