Definite Integral Calculator

Why F(b) − F(a) gives the area

Imagine an area function $A(t)$ that measures the area under $f$ from $a$ up to a moving point $t$. If you nudge $t$ forward by a tiny amount $h$, the extra area is a thin strip of height about $f(t)$ and width $h$, so $A(t+h) - A(t) \approx f(t)\,h$. Dividing by $h$ and letting it shrink shows that $A'(t) = f(t)$: the area function is an antiderivative of $f$.

Any two antiderivatives differ only by a constant, so $A(t) = F(t) + C$ for whichever antiderivative $F$ you found. Because $A(a) = 0$, the constant is $-F(a)$, and the full area is $A(b) = F(b) - F(a)$. This is also why the constant of integration never appears in a definite integral: it cancels in the subtraction.

Signed area and absolute area

When a graph dips below the axis, the integral subtracts that part. For $\int_0^{2\pi} \sin x\,dx$ the positive hump and the negative hump cancel exactly, and the answer is 0. That is the correct value of the integral, but it is not the area you would shade on a graph.

To find the total area, locate the zeros of the function inside the interval, integrate on each piece and add the absolute values of the pieces. The calculator accepts abs(), so entering abs(x-1) from 0 to 2 returns 1, the combined area of two triangles, while x-1 over the same interval returns 0.

Improper integrals: infinite limits and vertical asymptotes

An integral is improper when an endpoint is infinite or when the integrand becomes unbounded somewhere on the interval. The value is then defined as a limit. For $\int_1^{\infty} \frac{1}{x^2}\,dx$, the calculator finds $F(x) = -\frac1x$, takes the limit as $x \to \infty$, which is 0, subtracts $F(1) = -1$ and reports that the integral converges to 1.

The same idea handles asymptotes. $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges to 2 because $2\sqrt{x}$ has a finite limit at 0, while $\int_0^1 \frac1x\,dx$ diverges because $\ln x$ heads to $-\infty$. Both integrands blow up at zero; what decides the outcome is how fast. If the singularity is in the middle of the interval, the integral must be split there and each half must converge separately.

Over the whole real line both ends need a limit. The calculator evaluates $\int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx$ as $\arctan x$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ and returns exactly $\pi$.

When there is no formula: numerical integration

Some perfectly ordinary functions have antiderivatives that cannot be written with elementary functions. The bell-shaped $e^{-x^2}$, $\frac{\sin x}{x}$ and $\sqrt{1 + x^3}$ are standard examples. Their definite integrals still exist; they simply have to be computed numerically.

In that case the calculator uses adaptive Simpson's rule, which fits parabolas to small pieces of the curve and subdivides wherever the estimate is not yet accurate, aiming for an error below $10^{-10}$. For $\int_0^1 e^{-x^2}\,dx$ it reports approximately 0.7468241328, labelled as a numerical approximation so it is never confused with an exact value.

Useful properties that save work

Swapping the limits changes the sign, so $\int_1^0 x^2\,dx = -\frac13$. An integral over an interval of zero width is zero. For an odd function such as $x^3$ or $\sin x$, the integral over a symmetric interval $[-a, a]$ is zero, and for an even function it is twice the integral from 0 to $a$. Spotting symmetry before calculating can turn a messy problem into a one-line answer.

Integrals are also linear: constants factor out and sums split into separate integrals. Combined with the interval-splitting rule, these properties let you break a complicated region into pieces that are each easy to handle.