Integrals and the Area Under a Curve

An integral measures accumulated quantity

The definite integral ∫ₐᵇ f(x) dx measures the total accumulation of f between a and b. If f(x) is a rate — speed, rainfall per hour, dollars per item — then the integral of that rate over an interval is the total amount: total distance, total rainfall, total cost. Integration adds up infinitely many infinitesimal pieces, dx, each weighted by f(x).

The most direct picture is geometric: when f(x) is positive on [a, b], the integral equals the area enclosed by the curve, the x-axis, and the vertical lines x = a and x = b. Every application of integrals is some version of this idea — area first, accumulation in general.

Signed area: why area can be negative

When the curve dips below the x-axis, the integral counts that region as negative area. The integral is signed area: regions above the axis add, regions below subtract. So ∫₀^{2π} sin(x) dx = 0, because the positive hump from 0 to π and the negative trough from π to 2π have exactly equal areas and cancel.

This cancellation is a feature, not a bug: it reflects genuine physics. If velocity is positive on the first half of a trip and negative on the second, the integral gives displacement (net change in position), which can be zero even though the odometer ran up distance. If you want total area regardless of sign, integrate |f(x)| or integrate the positive and negative parts separately.

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus ties integrals and derivatives together as inverses. If F is an antiderivative of f — meaning F′(x) = f(x) — then ∫ₐᵇ f(x) dx = F(b) − F(a). Instead of approximating an area with thousands of rectangles, you evaluate a single function at two points and subtract.

This is why integrals are computed symbolically where possible: the antiderivative of 2*x is x^2, so ∫₀³ 2*x dx = 3² − 0² = 9, and you can verify this equals the area of a triangle with base 3 and height 6. The theorem turns the hard problem (adding up infinitely many slivers) into the easy problem (evaluating a function twice).

Area between two curves

Integrals also measure the area trapped between two curves. If g(x) ≤ h(x) on [a, b], the region between them has area ∫ₐᵇ (h(x) − g(x)) dx. You subtract the lower curve from the upper one, turning the gap into an ordinary area-under-a-curve problem.

For example, between x = 0 and x = 1 the line y = x lies above the curve y = x². The area between them is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6. A common mistake is forgetting that the curves can cross: where they swap roles, split the integral at the crossing points and integrate |h(x) − g(x)|, or you will let signed area cancel regions that should add.

Trying integrals in the calculator

Pick any expression below and use the calculator's integral tools to shade the area under the curve between two bounds. Watch how the shaded region flips sign when the curve crosses the axis — the tool reports signed area, so a symmetric wave like sin(x) over a full period nets to zero.

Then try the antiderivative relationship: plot f(x) and its integral-derived accumulation together. Where f is positive, the accumulated curve rises; where f is negative, it falls; where f is zero, it levels off. That connection — the derivative of the accumulation is the original function — is the Fundamental Theorem in visible form.

Key takeaways

• The definite integral ∫ₐᵇ f(x) dx measures accumulated quantity — geometrically, the area under the curve when f is positive.
• Integrals compute signed area: regions below the x-axis count as negative and can cancel regions above.
• The Fundamental Theorem of Calculus: ∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f — differentiation and integration undo each other.
• The area between two curves is ∫ₐᵇ (upper − lower) dx; split the integral wherever the curves cross.
• Displacement vs. distance: the integral of velocity gives net change; integrating the absolute value gives total distance traveled.

Frequently asked questions