Q.Evaluate the definite integral: $\int_4^5 e^x \ dx$

The key idea is that the integral of $e^x$ is itself, $e^x$, plus the constant of integration.

Step 1: Recall the antiderivative:

Step 2: Apply the Fundamental Theorem of Calculus to the limits $4$ and $5$:

$\int_4^5 e^x \, dx = \left[ e^x \right]_4^5 = e^5 - e^4$ …

The key to this problem is one of the most beautiful facts in calculus: the exponential function $e^x$ is its own derivative and its own antiderivative. No other function behaves this way (up to a constant factor). This means that when you integrate $e^x$, you don't need to worry about power rules, logarithms, or any special tricks — you just get $e^x$ back, plus the constant of integration for indefinite integrals.

For a definite integral, this property makes evaluation almost trivial: you find the antiderivative at the upper limit, subtract the antiderivative at the lower limit, and you're done.

Let's walk through it step by step.

1. Recall the fundamental theorem of calculus. For a continuous function $f(x)$ on $[a, b]$, if $F(x)$ is any antiderivative of $f(x)$, then

Here, $f(x) = e^x$.

2. Identify the antiderivative.

   Since $\frac{d}{dx}(e^x) = e^x$, it follows that $\int e^x \, dx = e^x + C$. So we can take $F(x) = e^x$.

3. Apply the limits.

   The integral from $4$ to $5$ is:

4. Simplify if desired. Factor out $e^4$:

This is a compact form, but $e^5 - e^4$ is perfectly acceptable as a final answer. …

Method: Definite integral of $e^{x}$ (self-integrating exponential)

Since $\int e^{x}\,dx=e^{x}$, the definite value is simply $e^{b}-e^{a}$.

Steps

Step 1: Recall the antiderivative. $\int e^{x}\,dx=e^{x}+C$.

Step 2: Bracket across the limits.

$\int_{a}^{b}e^{x}\,dx=\big[e^{x}\big]_{a}^{b}=e^{b}-e^{a}.$ …

Common Mistakes

Mistake 1: Writing $e^{b}-e^{a}=e^{\,b-a}$.

Why it's wrong: exponentials do not subtract by subtracting exponents; $e^{5}-e^{4}\neq e^{1}$. Correct approach: evaluate each term, then subtract (or factor as $e^{4}(e-1)$).

Mistake 2: "Integrating" $e^{x}$ to $\dfrac{e^{x}}{x}$ or $x e^{x-1}$. …