Imagine you're driving and your speedometer shows your speed at every moment. The integral of your speed over a time period gives you the total distance travelled. It's the accumulation of all those tiny little speed × time pieces added together.
Geometrically, the integral of a function over an interval is the area between the curve and the x-axis. If you sliced that area into infinitely many infinitely thin rectangles and added them all up, that's what an integral computes.
There are two kinds: an indefinite integral gives you a family of functions (the antiderivative), while a definite integral gives you a specific number — the accumulated total between two bounds.
Divide the area under a curve into \(n\) thin vertical rectangles. Each rectangle has width \(\Delta x\) and height \(f(x_i)\), so its area is \(f(x_i)\,\Delta x\). Add them all up: \(\sum f(x_i)\,\Delta x\). Now let \(n \to \infty\) so the rectangles become infinitely thin — the sum becomes an integral: \(\int f(x)\,dx\). This is the Riemann sum definition.
If the derivative asks "how fast is this changing?", the integral asks "given how fast it's changing, what's the total accumulated change?" They are inverse operations — which is exactly what the Fundamental Theorem of Calculus states. That's why we call the indefinite integral an antiderivative.
The definite integral is defined as the limit of Riemann sums:
Here \(x_i^*\) is a sample point in the \(i\)-th subinterval. As \(n \to \infty\) the rectangles become infinitely thin and the sum converges to the exact area.
| Power Rule | \(\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\) |
| Constant | \(\displaystyle\int c\,dx = cx + C\) |
| Exponential | \(\displaystyle\int e^x\,dx = e^x + C\) |
| Natural Log | \(\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C\) |
| Sine | \(\displaystyle\int \sin x\,dx = -\cos x + C\) |
| Cosine | \(\displaystyle\int \cos x\,dx = \sin x + C\) |
| Sum Rule | \(\displaystyle\int [f+g]\,dx = \int f\,dx + \int g\,dx\) |
| Constant Multiple | \(\displaystyle\int c\cdot f\,dx = c\int f\,dx\) |
\(\displaystyle\int e^u\,du = e^{x^2}+C\)
Area = 2 square units
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