A definite integral is an accumulation calculation over an interval. Sometimes it is literally “area under a curve,” and sometimes it is “net change” (like total distance from velocity). The workflow is simple: find an antiderivative and then evaluate it at the bounds.
Want your bounds and signs checked?
Use our ai math solver to see the antiderivative, the substitution (if any), and the final evaluation F(b) − F(a).
If you can find an antiderivative F of f, then the Fundamental Theorem of Calculus (FTC) tells you:
Two practical notes:
1) For definite integrals, there is no + C because it cancels in F(b) − F(a).
2) Use parentheses when plugging in bounds. Most mistakes are sign mistakes.
A definite integral is signed area: regions above the x-axis contribute positive area, and regions below contribute negative area. That is perfect for “net change” interpretations, but if the question explicitly asks for geometric area, you may need to split at x-intercepts and take absolute values.
Rule of thumb:
Problem: Evaluate ∫_0^2 (3x^2 − 4) dx.
First find an antiderivative:
Now apply FTC:
Getting 0 is not “wrong” by default: it means the positive and negative signed areas cancel over the interval.
Problem: Find the area between f(x) = x − 2 and the x-axis on the interval[0, 5].
The function crosses the x-axis where x − 2 = 0 → x = 2. On [0,2], the function is negative; on [2,5], it is positive. So split:
Compute each piece using antiderivatives:
Notice the difference: the signed integral ∫_0^5 (x − 2) dx would subtract the negative part; area adds it.
When you do u-sub in a definite integral, you must avoid ending with “u” but bounds in “x.” There are two safe options:
Do the u-sub to integrate, then rewrite the antiderivative back in terms of x, and only then plug in a and b.
Convert x-bounds to u-bounds using u = g(x). Then you integrate and evaluate entirely in u.
If you want a deeper conceptual picture (how sums turn into areas and why FTC works), OpenStax has a clear explanation in its section on the definite integral. You don’t need every detail to compute problems, but the interpretation helps you catch sign and “area vs net” mistakes.
Problem: Evaluate ∫_0^1 2x (x^2 + 1)^4 dx.
Let u = x^2 + 1, so du = 2x dx. Now convert bounds: when x = 0, u = 1; when x = 1, u = 2.
This is the cleanest way to keep variables consistent. It also makes the final evaluation faster.
Problem: A particle has velocity v(t) = 3t − 2 (meters/second). Find the displacement from t = 0 to t = 4.
Displacement is the integral of velocity:
Notice how the bounds act like “start time” and “end time.” If the velocity went negative on part of the interval, the integral would subtract that part (net displacement). If you were asked for total distance, you would split where v(t) = 0.
A common “definite integral” application is average value. The average value of f(x) on[a,b] is:
The structure mirrors “average = total / length.” The integral is the total accumulation, and (b − a)is the interval length.
Always write F(b) − F(a) with brackets: [F(b)] − [F(a)]. One missing bracket can flip the sign.
If the question asks for area, split at zeros (where the curve crosses the axis) or use absolute value.
Don’t finish with u but evaluate using x-bounds. Either change bounds to u, or substitute back to x.
If x is time and f(x) is velocity, the integral’s units are distance. Units help sanity-check your answer.
For #4, remember: “area” means you should split at the x-intercepts (here, where x^2 − 1 = 0).
Paste a definite integral and get the antiderivative + evaluation with correct signs.
Try AI Math Solver →