Power rule on a parabola
\(\int_0^2 x^2\,dx = \frac{x^3}{3}\Big|_0^2 = \frac{8}{3} \approx 2.667\)
Integral Calculator
Calculate indefinite and definite integrals with step-by-step solutions
Explore the definite integral
Type a function in x or pick a preset, then drag the lower (a) and upper (b) bounds. The shaded region under f(x) is the signed area, equal to F(b) − F(a).
0.00
-10.0010.00
2.00
-10.0010.00
Polynomials, sin/cos/exp with linear arguments, and 1/x are supported. Use x as the variable.
Common integrals
Predict what will happen
What happens to ∫ₐᵇ f(x) dx if you swap a and b?
Try setting a = 2 and b = 0 with the parabola preset, then a = 0 and b = 2.
Signed area
The integral is the SIGNED area between the curve and the x-axis: regions below contribute negatively. If f(x) is always positive on [a, b], the integral equals the geometric area.
Common mistake
Don't forget the constant of integration C in indefinite integrals. For definite integrals, the C cancels — F(b) − F(a) is the same regardless of C.
Why it works (FTC)
The Fundamental Theorem of Calculus connects differentiation and integration: if F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a). Integration accumulates change; differentiation measures the rate of change.
Results
Final Answer
\(\int_{0}^{2} x ^ 2 \,dx = 2.666667\)
Step-by-step Solution
- Definite integral: \(\int_{0}^{2} x ^ 2 \,dx\)
- Find the antiderivative term by term:
- ∫ \(x ^ 2\) dx = \(x ^ 3 / 3\)
- Antiderivative: \(F(x) = x ^ 3 / 3 + C\)
- Apply the Fundamental Theorem of Calculus: \(F(2) - F(0) = 2.666667 - (0) = 2.666667\)
Indefinite antiderivative: \(\int x ^ 2 \,dx = x ^ 3 / 3 + C\)
Theory & Formula
Integration is the reverse process of differentiation. It finds the antiderivative of a function, representing the area under a curve.
Fundamental Theorem of Calculus: \(\int_a^b f(x)\,dx = F(b) - F(a)\)
If F is any antiderivative of f on [a, b], then ∫ₐᵇ f(x) dx = F(b) − F(a). Differentiation undoes integration; integration accumulates the change measured by differentiation.
\(\int_a^b f(x)\,dx = F(b) - F(a)\)
Worked Examples
Sine over half a period
\(\int_0^{\pi} \sin(x)\,dx = -\cos(x)\Big|_0^{\pi} = 2\)