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Definite Integral

Definite Integral

An integral which is evaluated over an interval. A definite integral is written Definite integral notation: integral from a to b of f(x)dx, where a and b are the lower and upper bounds of integration.. Definite integrals are used to find the area between the graph of a function and the x-axis. There are many other applications.

Formally, a definite integral is the limit of a Riemann sum as the norm of the partition approaches zero. That is, Definite integral formula: integral from a to b of f(x)dx equals limit as n→∞ of sum from k=1 to n of f(c_k)(x_k − x_{k−1}).

 

Example: definite integral from 2 to 5 of x² dx = (1/3)x³ evaluated from 2 to 5 = (1/3)(5)³ − (1/3)(2)³ = 117/3 = 39

 

 

See also

Fundamental Theorem of Calculus, integration methods, integral rules, work, volume by parallel cross-sections, disk method, cylindrical shell method

Key Formula

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)
Where:
  • aa = The lower limit of integration (left endpoint of the interval)
  • bb = The upper limit of integration (right endpoint of the interval)
  • f(x)f(x) = The integrand — the function being integrated
  • F(x)F(x) = Any antiderivative of f(x), meaning F'(x) = f(x)
  • dxdx = Indicates the variable of integration is x

Worked Example

Problem: Evaluate the definite integral ∫142x dx\int_1^4 2x\,dx.
Step 1: Find an antiderivative of the integrand f(x)=2xf(x) = 2x.
F(x)=x2F(x) = x^2
Step 2: Apply the Fundamental Theorem of Calculus: evaluate F(x)F(x) at the upper limit b=4b = 4 and the lower limit a=1a = 1.
∫142x dx=F(4)−F(1)\int_1^4 2x\,dx = F(4) - F(1)
Step 3: Substitute the limits into the antiderivative.
F(4)=42=16F(1)=12=1F(4) = 4^2 = 16 \qquad F(1) = 1^2 = 1
Step 4: Subtract to get the final value.
16−1=1516 - 1 = 15
Answer: ∫142x dx=15\int_1^4 2x\,dx = 15. This means the net signed area between f(x)=2xf(x) = 2x and the xx-axis from x=1x = 1 to x=4x = 4 is 15 square units.

Another Example

This example uses a trigonometric function instead of a polynomial, showing that the same Fundamental Theorem approach works regardless of the type of integrand. It also involves careful sign handling with cosine values.

Problem: Evaluate ∫0πsin⁡(x) dx\int_0^{\pi} \sin(x)\,dx.
Step 1: Identify an antiderivative of sin⁡(x)\sin(x).
F(x)=−cos⁡(x)F(x) = -\cos(x)
Step 2: Evaluate F(x)F(x) at the upper limit b=πb = \pi.
F(π)=−cos⁡(π)=−(−1)=1F(\pi) = -\cos(\pi) = -(-1) = 1
Step 3: Evaluate F(x)F(x) at the lower limit a=0a = 0.
F(0)=−cos⁡(0)=−(1)=−1F(0) = -\cos(0) = -(1) = -1
Step 4: Subtract to find the definite integral.
F(π)−F(0)=1−(−1)=2F(\pi) - F(0) = 1 - (-1) = 2
Answer: ∫0πsin⁡(x) dx=2\int_0^{\pi} \sin(x)\,dx = 2.

Frequently Asked Questions

What is the difference between a definite integral and an indefinite integral?
A definite integral has specific limits of integration aa and bb, and it produces a single numerical value representing the net signed area under a curve. An indefinite integral has no limits and produces a family of antiderivatives, written as F(x)+CF(x) + C, where CC is an arbitrary constant.
Can a definite integral be negative?
Yes. A definite integral measures net signed area. When the function lies below the xx-axis, it contributes negative area. If the function spends more of the interval below the axis than above it, the definite integral will be negative. For example, ∫0π−sin⁡(x) dx=−2\int_0^{\pi} -\sin(x)\,dx = -2.
What happens if you swap the limits of a definite integral?
Swapping the limits reverses the sign of the integral. Formally, ∫baf(x) dx=−∫abf(x) dx\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx. This follows directly from the Fundamental Theorem: F(a)−F(b)F(a) - F(b) is the negative of F(b)−F(a)F(b) - F(a).

Definite Integral vs. Indefinite Integral

Definite IntegralIndefinite Integral
Notation∫abf(x) dx\int_a^b f(x)\,dx∫f(x) dx\int f(x)\,dx
Limits of integrationHas specific bounds aa and bbNo bounds
ResultA single numberA function F(x)+CF(x) + C
Constant of integrationNot needed — it cancels during subtractionRequired (+C+ C)
InterpretationNet signed area under the curve on [a,b][a, b]General antiderivative of f(x)f(x)

Why It Matters

Definite integrals appear throughout AP Calculus, college calculus, physics, and engineering. You use them to compute areas, volumes of solids, total displacement, work done by a force, and accumulated quantities. Mastering the definite integral connects the concept of antiderivatives (from differential calculus) to real-world measurement, making it one of the most applied ideas in all of mathematics.

Common Mistakes

Mistake: Forgetting to evaluate the antiderivative at both limits and subtract. Some students find F(b)F(b) but forget to subtract F(a)F(a), or they evaluate only at one endpoint.
Correction: Always compute F(b)−F(a)F(b) - F(a). A helpful notation is to write [F(x)]ab\bigl[F(x)\bigr]_a^b immediately after finding the antiderivative, which reminds you to plug in both limits and subtract.
Mistake: Confusing total area with the value of the definite integral. When a function dips below the xx-axis, the definite integral subtracts that region, but total area requires you to integrate the absolute value ∣f(x)∣|f(x)|.
Correction: If a problem asks for total area, split the integral at the zeros of f(x)f(x) and take the absolute value of each piece, or integrate ∣f(x)∣|f(x)| directly.

Related Terms