Evaluating Definite Integrals by Endpoint Substitution

From an Antiderivative to a Definite Integral

A definite integral has a numerical value: it measures the net, or signed, accumulation of a function across an interval. When the integrand has a known antiderivative, the Fundamental Theorem of Calculus gives a direct way to find that value. Instead of constructing a limit of sums, we evaluate one function at two endpoints and subtract.

The order of that subtraction matters. For an integral from $a$ to $b$, evaluate the antiderivative at the upper endpoint $b$ first, then subtract its value at the lower endpoint $a$. This procedure is called endpoint substitution.

Theorem: If $f$ is continuous on $[a,b]$, and $F$ is an antiderivative of $f$ on an interval containing $[a,b]$, so that $F'(x)=f(x)$, then $$ \int_a^b f(x)\,dx=F(b)-F(a). $$

Continuity of $f$ on the interval is a standard sufficient condition in AP Calculus. The function $F$ used in the formula must actually be an antiderivative of $f$ throughout the interval, not just a function whose derivative matches at the endpoints.

Notation for Substituting the Endpoints

The expression $F(b)-F(a)$ is often written more compactly as $[F(x)]_a^b$. The upper endpoint is written above the lower endpoint, and the evaluation always means upper value minus lower value:

$$ [F(x)]_a^b=F(b)-F(a). $$

For instance, if an antiderivative is $F(x)=x^3$, then $[x^3]_1^2$ means $2^3-1^3$, not $1^3-2^3$. The variable $x$ in the bracket is a placeholder: substitute each endpoint into the expression for $F$, then carry out the subtraction.

In an indefinite integral, one usually writes a constant of integration, $+C$, because there are many antiderivatives of a function. In a definite integral, a single antiderivative can be used without writing $+C$. If we did include the constant, it would cancel:

$$ (F(b)+C)-(F(a)+C)=F(b)-F(a). $$

This cancellation is one reason endpoint substitution gives a well-defined value even though an antiderivative is not unique. More precisely, the formula works because of the Fundamental Theorem; the cancellation explains why the arbitrary constant in an antiderivative cannot affect the result.

A Reliable Evaluation Process

For a definite integral that can be evaluated by a known antiderivative, organize the work so that the antiderivative and both endpoint values remain visible. This makes the sign and arithmetic easier to check.

Find an antiderivative.
Choose $F$ so that $F'(x)=f(x)$ on the interval of integration.

Substitute the upper endpoint.
Compute $F(b)$, preserving parentheses when substituting a negative or complicated value.

Substitute the lower endpoint.
Compute $F(a)$ separately before subtracting it.

Subtract and check.
Form $F(b)-F(a)$, simplify, and differentiate $F$ to verify that it returns the integrand.

Worked Examples

Worked Example: A Polynomial Integrand

Evaluate $\int_0^2 (3x^2-4x+1)\,dx$.

First find an antiderivative, integrating each term:

\int (3x^2-4x+1)\,dx =x^3-2x^2+x.

Set $F(x)=x^3-2x^2+x$. Check the derivative before using it:

$$ F'(x)=3x^2-4x+1. $$

Now substitute the upper endpoint $2$:

$$ F(2)=2^3-2(2^2)+2=8-8+2=2. $$

Substitute the lower endpoint $0$:

$$ F(0)=0^3-2(0^2)+0=0. $$

Subtract lower from upper:

\int_0^2 (3x^2-4x+1)\,dx =F(2)-F(0)=2-0=2.

The result is $2$. The derivative check confirms that the chosen antiderivative matches the integrand, and the endpoint values give the stated difference.

Worked Example: A Trigonometric Integrand

Evaluate $\int_0^{\pi/2}(2\sin x+\cos x)\,dx$.

Using $\int \sin x\,dx=-\cos x$ and $\int \cos x\,dx=\sin x$, take

F(x)=-2\cos x+\sin x.

Differentiate to check:

F'(x)=2\sin x+\cos x.

At the upper endpoint $x=\pi/2$, use $\cos(\pi/2)=0$ and $\sin(\pi/2)=1$:

F\left(\frac{\pi}{2}\right) =-2\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right) =-2(0)+1=1.

At the lower endpoint $x=0$, use $\cos(0)=1$ and $\sin(0)=0$:

F(0)=-2\cos(0)+\sin(0)=-2(1)+0=-2.

Therefore, upper value minus lower value is

\int_0^{\pi/2}(2\sin x+\cos x)\,dx =1-(-2)=3.

The answer is $3$. In particular, subtracting the lower value means subtracting $-2$, which increases the result.

Worked Example: An Exponential Integrand

Evaluate $\int_0^{\ln 3}(e^x+2)\,dx$.

An antiderivative is

$$ F(x)=e^x+2x, $$

because

$$ F'(x)=e^x+2. $$

At the upper endpoint $x=\ln 3$, use $e^{\ln 3}=3$:

F(\ln 3)=e^{\ln 3}+2\ln 3=3+2\ln 3.

At the lower endpoint $x=0$, use $e^0=1$:

$$ F(0)=e^0+2(0)=1. $$

Now subtract:

\int_0^{\ln 3}(e^x+2)\,dx =F(\ln 3)-F(0) =(3+2\ln 3)-1 =2+2\ln 3.

The exact answer is $2+2\ln 3$. No decimal approximation is needed; keeping the logarithm gives an exact value. Differentiating $e^x+2x$ verifies the antiderivative used in the calculation.

Why Endpoint Substitution Works

The result is the evaluation form of the Fundamental Theorem of Calculus. A brief way to see why the endpoint difference is natural is to define the accumulation function $G(x)=\int_a^x f(t)\,dt$. When $f$ is continuous, the Fundamental Theorem gives $G'(x)=f(x)$. If $F$ is also an antiderivative of $f$, then the difference $G-F$ has derivative zero on the interval:

$$ (G-F)'(x)=G'(x)-F'(x)=f(x)-f(x)=0. $$

By the Mean Value Theorem, a differentiable function whose derivative is zero throughout an interval is constant there. Thus $G(b)-F(b)=G(a)-F(a)$. Since $G(a)=\int_a^a f(t)\,dt=0$, rearranging gives

\int_a^b f(t)\,dt=G(b)=F(b)-F(a).

This explains the endpoint rule: the integral records the change in an accumulation function, and an antiderivative has the same rate of change. The difference between its endpoint values therefore gives the net accumulation.

Common Pitfall: Reversing the Subtraction

A frequent error is to evaluate the lower endpoint first and subtract the upper endpoint. For example, consider $\int_0^2 2x\,dx$. An antiderivative is $F(x)=x^2$, since $F'(x)=2x$. The incorrect order gives

$$ F(0)-F(2)=0^2-2^2=0-4=-4. $$

That is not the value of the integral. The required order is upper minus lower:

\int_0^2 2x\,dx=F(2)-F(0)=2^2-0^2=4.

The correction is not to take an absolute value after the calculation; it is to apply the endpoint order correctly. Also remember that a definite integral is a signed accumulation. If the integrand is below the horizontal axis over part of the interval, that part contributes negatively. The integral need not equal the total geometric area between the graph and the axis.

Checks Before You Finish

A correct endpoint calculation depends on both the antiderivative and the evaluation. Before finalizing an answer, verify that differentiating $F$ recovers the integrand, that the upper endpoint appears first, and that each endpoint has been substituted into the entire expression. Parentheses are especially useful when the endpoint is negative or when the antiderivative contains several terms.

Exact values should generally be retained when they are simple expressions such as $2+2\ln 3$ or $3$. If a problem asks for a decimal, evaluate the exact expression first and round only at the final step. For an integral with units, the result has the units of the integrand multiplied by the units of the variable of integration.

Key takeaway: If $F'(x)=f(x)$ on the interval, then evaluate $\int_a^b f(x)\,dx$ by calculating $F(b)-F(a)$. The order is always upper endpoint minus lower endpoint.

Check Your Understanding

For each question, show the antiderivative and the two endpoint values before simplifying.

Evaluate $\int_1^3 2x\,dx$, and check your antiderivative by differentiating.
Evaluate $\int_0^\pi \sin x\,dx$. What are the upper and lower endpoint values of an antiderivative?
Evaluate $\int_0^{\ln 2} e^x\,dx$ exactly.
For an antiderivative $F$, what does $[F(x)]_a^b$ mean?
Explain why $F(a)-F(b)$ uses the wrong order for $\int_a^b f(x)\,dx$.

Evaluating Definite Integrals by Endpoint Substitution

What You'll Learn

From an Antiderivative to a Definite Integral

Notation for Substituting the Endpoints

A Reliable Evaluation Process