Q.Find an anti derivative (or integral) of the function $e^{2x}$ by the method of inspection

The key idea is Antiderivative By Inspection: we ask which function, when differentiated, gives $e^{2x}$.

1. Recall that $\frac{d}{dx} e^{2x} = 2e^{2x}$ (by the chain rule).
2. To get exactly $e^{2x}$, we need to cancel the factor of $2$ that appears upon differentiation.
3. Differentiate $\frac{1}{2} e^{2x}$: $\frac{d}{dx} \left( \frac{1}{2} e^{2x} \right) = \frac{1}{2} \cdot 2 e^{2x} = e^{2x}$.

The method of inspection for finding antiderivatives is really just reverse differentiation. You ask yourself: "What function, when differentiated, gives me this?" It's like looking at a finished jigsaw puzzle and figuring out what picture was on the box.

For exponential functions, the key fact is that the derivative of $e^{ax}$ is $ae^{ax}$. The function $e^{2x}$ is almost its own derivative, except for that extra factor of 2 that appears when we differentiate. So we need to "undo" that factor.

1. Start with the target. We want a function $F(x)$ such that $F'(x) = e^{2x}$.

2. Think about the derivative of $e^{2x}$. If we differentiate $e^{2x}$, we get $2e^{2x}$. That's close, but it's off by a factor of 2.

3. Adjust the constant. Since differentiation is linear, if we multiply $e^{2x}$ by $\frac{1}{2}$, the derivative will also be multiplied by $\frac{1}{2}$. So:

4. Check your work. Differentiate $\frac{1}{2}e^{2x}$: the derivative of $e^{2x}$ is $2e^{2x}$, multiplied by $\frac{1}{2}$ gives exactly $e^{2x}$. It works.

5. Don't forget the constant. Every antiderivative is actually a family of functions. Since the derivative of any constant is zero, we can add any constant $C$ to our answer and it will still differentiate to $e^{2x}$.

So the antiderivative (or indefinite integral) of $e^{2x}$ is $\frac{1}{2}e^{2x} + C$, where $C$ is any constant.

Method: Antiderivative of an Exponential by Inspection

Use this whenever the integrand is $e^{ax}$ (or a constant times it): the antiderivative is the same exponential with the constant rescaled.

Steps

Step 1: Recall the derivative rule for $e^{ax}$.

By the chain rule, $\dfrac{d}{dx}e^{ax}=a\,e^{ax}$. So $e^{ax}$ is almost its own antiderivative — it is off only by the factor $a$ that appears when you differentiate.

Step 2: Cancel the stray factor.

To undo that factor of $a$, divide by $a$:

Step 3: Verify.

Differentiate the candidate: $\dfrac{d}{dx}\!\left(\dfrac1a e^{ax}\right)=\dfrac1a\cdot a\,e^{ax}=e^{ax}$, which matches the integrand. Attach $+C$.

This general pattern works for any linear exponent $ax+b$ as well: $\int e^{ax+b}\,dx=\dfrac1a e^{ax+b}+C$.

Common Mistakes

Mistake 1: Writing $\int e^{2x}\,dx = e^{2x} + C$.

Why it's wrong: this ignores the chain-rule factor — $\frac{d}{dx}e^{2x}=2e^{2x}$, so $e^{2x}$ is not its own antiderivative. Correct approach: divide by the coefficient of $x$, giving $\tfrac12 e^{2x}+C$.

Mistake 2: Multiplying by the coefficient instead of dividing.

Why it's wrong: students sometimes write $2e^{2x}+C$, but that would differentiate to $4e^{2x}$. Correct approach: when integrating you divide by $a$, not multiply.

Mistake 3: Forgetting $+C$.

Why it's wrong: the antiderivative is only determined up to a constant. Correct approach: always include $+C$.