Introduction
A definite integral is often evaluated by finding an antiderivative and substituting the limits of integration. Bracket notation makes that substitution compact, but compact notation is useful only if its order is read correctly: the value at the upper endpoint comes first, and the value at the lower endpoint is subtracted.
This tutorial focuses on writing and interpreting that last step. The Fundamental Theorem of Calculus supplies the evaluation rule; the bracket records exactly how to use it. In particular, grouping each endpoint value with parentheses helps prevent sign errors when the lower-endpoint value is negative or contains several terms.
The endpoint-evaluation notation
Suppose \(F\) is an antiderivative of \(f\) on an interval containing \(a\) and \(b\). This means that \(F'(x)=f(x)\) throughout that interval. When \(f\) is continuous there, the Fundamental Theorem of Calculus gives the definite integral as the difference between the endpoint values of \(F\).
Thus the endpoint-evaluation form of the Fundamental Theorem of Calculus is \[ \int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a), \] provided \(f\) is continuous on the interval from \(a\) to \(b\) and \(F'=f\) on that interval. The symbol \(x\) inside the bracket is the input of \(F\); substitute each endpoint for it. The limits \(a\) and \(b\) are not added to the antiderivative, and they are not inputs to the integrand after the integration has been performed.
The notation is a compact way to write two substitutions and one subtraction. If the endpoint values are complicated, it is helpful to expand the bracket explicitly before simplifying: \[ [F(x)]_a^b=F(b)-F(a). \] The parentheses around \(F(a)\) matter. They tell us to subtract the entire lower-endpoint value, including all its terms and signs.
A reliable substitution routine
For a definite integral with a known antiderivative, keep the calculation in a consistent order:
Choose \(F\) with \(F'(x)=f(x)\) on the interval of integration.
Place the integration variable in the antiderivative and put the lower and upper limits beneath and above the bracket: \([F(x)]_a^b\).
Rewrite the bracket as \(F(b)-F(a)\), keeping the lower value in parentheses.
Compute the two endpoint values, subtract in the stated order, and confirm that the antiderivative differentiates to the integrand.
This routine separates the mathematical rule from the arithmetic. First establish what the notation means; then evaluate each endpoint. In particular, do not change the order merely because one endpoint value looks simpler or has a minus sign.
Worked Example: A polynomial on a numerical interval
Worked Example: A polynomial on a numerical interval
Evaluate \(\displaystyle \int_{-2}^{1}(2x+3)\,dx\) using bracket notation.
An antiderivative of \(2x+3\) is \(F(x)=x^2+3x\), since \[ F'(x)=2x+3. \] The integrand is a polynomial, so it is continuous on \([-2,1]\), and the Fundamental Theorem applies.
Write the endpoint evaluation: \[ \int_{-2}^{1}(2x+3)\,dx=[x^2+3x]_{-2}^{1}. \]
Substitute the upper endpoint first and the lower endpoint second: \[ [x^2+3x]_{-2}^{1} =(1^2+3(1))-((-2)^2+3(-2)). \]
Evaluate the two grouped expressions: \[ (1+3)-(4-6)=4-(-2)=6. \] Therefore, \[ \int_{-2}^{1}(2x+3)\,dx=6. \]
Check: Differentiating \(x^2+3x\) gives \(2x+3\), the required integrand. The endpoint values are \(F(1)=4\) and \(F(-2)=-2\), so \(F(1)-F(-2)=4-(-2)=6\), agreeing with the result.
Worked Example: Keeping a multi-term lower value together
Worked Example: Keeping a multi-term lower value together
Evaluate \(\displaystyle \int_0^2(3x^2-4x+1)\,dx\).
An antiderivative is \[ F(x)=x^3-2x^2+x, \] because \[ F'(x)=3x^2-4x+1. \] The integrand is continuous on \([0,2]\), so endpoint evaluation is valid.
Use the bracket and expand it as upper minus lower: \[ \int_0^2(3x^2-4x+1)\,dx =[x^3-2x^2+x]_0^2 =(2^3-2(2^2)+2)-(0^3-2(0^2)+0). \]
The upper-endpoint value is \[ 2^3-2(2^2)+2=8-8+2=2. \] The lower-endpoint value is \[ 0^3-2(0^2)+0=0. \] Thus \[ [x^3-2x^2+x]_0^2=2-0=2. \]
Check: The derivative of \(x^3-2x^2+x\) is \(3x^2-4x+1\). Substitution gives \(F(2)=2\) and \(F(0)=0\), so \(F(2)-F(0)=2\), as calculated.
Worked Example: Evaluating a symbolic pair of endpoints
Worked Example: Evaluating a symbolic pair of endpoints
Suppose \(a\) and \(b\) lie in an interval on which the integrand is continuous. Evaluate \(\displaystyle \int_a^b(2x+3)\,dx\) in terms of \(a\) and \(b\).
As before, take \(F(x)=x^2+3x\), for which \(F'(x)=2x+3\). Apply the endpoint notation without replacing \(a\) and \(b\) by numbers: \[ \int_a^b(2x+3)\,dx=[x^2+3x]_a^b. \]
Now substitute the upper endpoint \(b\), then subtract the value at the lower endpoint \(a\): \[ [x^2+3x]_a^b =(b^2+3b)-(a^2+3a). \] The result can also be written \[ b^2+3b-a^2-3a. \] Therefore, \[ \int_a^b(2x+3)\,dx=b^2+3b-a^2-3a. \]
Check: Differentiating \(x^2+3x\) gives \(2x+3\). The bracket has produced precisely \(F(b)-F(a)\), namely \(b^2+3b-(a^2+3a)\); distributing the minus sign gives the displayed expanded form. No numerical endpoint values are needed for this symbolic evaluation.
Worked Example: A trigonometric antiderivative
Worked Example: A trigonometric antiderivative
Evaluate \(\displaystyle \int_0^{\pi/2}\cos x\,dx\).
An antiderivative is \(F(x)=\sin x\), since \(\frac{d}{dx}\sin x=\cos x\). The cosine function is continuous on \([0,\pi/2]\), so the Fundamental Theorem applies.
Write and evaluate the bracket: \[ \int_0^{\pi/2}\cos x\,dx =[\sin x]_0^{\pi/2} =\sin\left(\frac{\pi}{2}\right)-\sin(0). \] Since \(\sin(\pi/2)=1\) and \(\sin(0)=0\), \[ [\sin x]_0^{\pi/2}=1-0=1. \] Hence \[ \int_0^{\pi/2}\cos x\,dx=1. \]
Check: The derivative of \(\sin x\) is \(\cos x\), and the endpoint values give \(F(\pi/2)-F(0)=1-0=1\), confirming the evaluation.
Why the bracket gives the integral
The bracket notation does not introduce a new integration rule. It abbreviates the endpoint difference in the Fundamental Theorem of Calculus. If \(F'=f\) and \(f\) is continuous on \([a,b]\), that theorem states that the accumulated integral from \(a\) to \(b\) is the change in \(F\) between those endpoints: \[ \int_a^b f(x)\,dx=F(b)-F(a). \] Writing \([F(x)]_a^b\) for the right-hand side is valid because the bracket is defined to mean exactly that difference.
The order is essential to the notation: a change from the starting endpoint to the ending endpoint is represented by ending value minus starting value. The notation therefore preserves the order of the definite integral’s displayed limits. It also makes a long expression shorter without changing its meaning, as long as the lower value remains subtracted in full.
Common mistake: subtracting in the wrong order
A frequent error is to calculate the lower-endpoint value first and subtract the upper-endpoint value from it. For the first example, this would give \[ F(-2)-F(1)=(-2)-4=-6. \] That is not what \([F(x)]_{-2}^{1}\) means. The bracket places \(1\) at the top, so the correct calculation is \[ [F(x)]_{-2}^{1}=F(1)-F(-2)=4-(-2)=6. \]
A closely related error is to omit parentheses around the lower-endpoint expression. For instance, \[ (b^2+3b)-(a^2+3a) \] must be expanded as \[ b^2+3b-a^2-3a, \] not \(b^2+3b-a^2+3a\). The minus sign applies to every term in the lower value. Keeping the entire lower substitution in parentheses until after the subtraction is performed avoids this mistake.
These checks are useful whenever an endpoint value is negative or contains multiple terms. Before simplifying, write the bracket explicitly as upper value minus lower value. Then evaluate the grouped expressions and distribute the subtraction only if needed.
Check Your Understanding
For each question, keep the upper-minus-lower order visible before simplifying.
- What does \([F(x)]_a^b\) mean in terms of \(F(a)\) and \(F(b)\)?
- Evaluate \([x^2-x]_1^3\), writing the two endpoint values in parentheses before subtracting.
- Find an antiderivative of \(4x+1\), then use bracket notation to express \(\displaystyle\int_0^c(4x+1)\,dx\) in terms of \(c\).
- For a continuous function \(f\) with antiderivative \(F\) on \([a,b]\), what conditions allow the endpoint-evaluation rule to be used?
- A student writes \([x^2+2x]_{-1}^{2}=F(-1)-F(2)\), where \(F(x)=x^2+2x\). Identify the order error and give the correct expression.
