MathematicsCalculusA-Level

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Area Under Curve

Definite integral calculation.

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A = \int_{a}^{b} f (x) d x == F (b) - F (a)

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Core idea

Overview

This formula represents the Second Fundamental Theorem of Calculus, which provides a computational method for evaluating definite integrals. It defines the net area under a curve as the difference between the values of the function's antiderivative evaluated at the upper and lower limits of integration.

When to use: Use this formula when calculating the accumulated change of a continuous function over a specific interval [a, b]. It is applicable whenever an antiderivative F(x) can be identified for the integrand f(x) such that F'(x) = f(x).

Why it matters: This relationship is the foundation of integral calculus, allowing scientists to solve complex problems in physics, engineering, and economics. It turns the geometric problem of finding areas into a straightforward algebraic calculation of evaluation.

Symbols

Variables

A = Area, F(b) = Upper Limit Val, F(a) = Lower Limit Val

A

Area

u ni t s^{2}

F (b)

Upper Limit Val

V a r iab l e

F (a)

Lower Limit Val

V a r iab l e

Walkthrough

Derivation

Understanding Area Under a Curve

A definite integral gives the signed area between a curve and the x-axis over an interval.

f(x) is continuous on [a, b].
Areas below the x-axis contribute negative values to the integral.

Write the Definite Integral:

Integrate from a to b to accumulate signed area.

Area (signed) = \int_{a}^{b} f (x) d x

Use the Fundamental Theorem of Calculus:

Find an antiderivative F(x), then substitute the limits.

\int_{a}^{b} f (x) d x = F (b) - F (a)

Note: If you want total geometric area, split at x-axis crossings and use absolute values.

Result

\int_{a}^{b} f (x) d x = F (b) - F (a)

Source: AQA A-Level Mathematics — Pure (Integration)

Visual intuition

Graph

The graph depicts a continuous function $f(x)$ plotted on a Cartesian plane, where the area under the curve is represented by the definite integral between two points $a$ and $b$. The shaded region bound by the curve, the x-axis, and the vertical lines $x=a$ and $x=b$ visually demonstrates the accumulation of quantities, such as distance or total growth. This geometric interpretation is central to the Fundamental Theorem of Calculus, linking the integration of the function to the net area bounded by the curve.

Graph type: polynomial

Why it behaves this way

Intuition

Imagine slicing the region under the curve f(x) into infinitely thin vertical rectangles, each with height f(x) and width dx, then summing the areas of all these slices from x=a to x=b to find the total area.

The net accumulated quantity or total change of the function f(x) over the interval [a, b].

This is the total 'amount' that has built up or changed from the starting point 'a' to the ending point 'b', as dictated by the function f(x).

f(x)

The instantaneous rate or value of the quantity being accumulated at a specific point x.

This represents the 'height' of the curve at any given x, indicating how much is being added (or subtracted) at that precise moment.

An infinitesimally small increment along the independent variable x.

This is the 'width' of an infinitely thin slice or interval over which f(x) is considered constant for the purpose of summation.

\int_{a}^{b} ... d x

The operation of definite integration, performing a continuous summation of f(x) multiplied by dx over the interval [a, b].

It's the process of adding up an infinite number of infinitesimally small contributions (f(x) * dx) from x=a to x=b.

F(b) - F(a)

The net change in the antiderivative F(x) from the lower limit 'a' to the upper limit 'b'.

This is the total accumulated amount at the end point 'b' minus the total accumulated amount at the start point 'a', directly yielding the total change over the interval.

Signs and relationships

F(b) - F(a): The subtraction calculates the net change in the accumulated quantity F(x) between the upper limit b and the lower limit a. A positive result indicates a net increase in the accumulated quantity, while a negative result

Free study cues

Insight

Canonical usage

This equation is used to determine an accumulated quantity, where the unit of the result 'A' is the product of the unit of the function 'f(x)' and the unit of the integration variable 'x'.

Common confusion

A common mistake is assuming the result 'A' always represents a geometric area (e.g., in m2 or cm2), rather than a physical quantity whose units are derived from the product of the integrand's units and the integration

Unit systems

$A$ Unit of f(x) × Unit of x · Represents the accumulated quantity or 'net area' under the curve. Its dimension is the product of the dimensions of the integrand f(x) and the independent variable x.

$f (x)$ Any valid unit · The integrand function. Its units, when multiplied by the units of x, determine the dimension of the accumulated quantity.

$x$ Any valid unit · The independent variable of integration. Its units define the 'width' of the infinitesimal elements summed by the integral.

$d x$ Unit of x · The differential of x, carrying the same dimension as x.

$a, b$ Unit of x · The lower and upper limits of integration, respectively. They must have the same dimension as the independent variable x.

$F (x)$ Unit of f(x) × Unit of x · The antiderivative of f(x). Its dimension must be consistent with the dimension of the definite integral A.

One free problem

Practice Problem

A particle moves along a path where the antiderivative of its velocity function represents its position. If the position at the end of the journey (Fb) is 50 meters and the position at the start (Fa) is 15 meters, calculate the total displacement (A) representing the area under the velocity curve.

Upper Limit Val50

Lower Limit Val15

Solve for: $A$

Hint: Subtract the initial antiderivative value from the final antiderivative value.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Distance travelled given velocity graph.

Study smarter

Tips

Always verify that the function is continuous over the entire interval [a, b].
Pay close attention to signs when subtracting the lower bound value from the upper bound value.
Identify the antiderivative accurately before substituting the boundary values.

Avoid these traps

Common Mistakes

Order of subtraction (F(a)-F(b)).
Forgetting to integrate first.

Common questions

Frequently Asked Questions

A definite integral gives the signed area between a curve and the x-axis over an interval.

Use this formula when calculating the accumulated change of a continuous function over a specific interval [a, b]. It is applicable whenever an antiderivative F(x) can be identified for the integrand f(x) such that F'(x) = f(x).

This relationship is the foundation of integral calculus, allowing scientists to solve complex problems in physics, engineering, and economics. It turns the geometric problem of finding areas into a straightforward algebraic calculation of evaluation.

Order of subtraction (F(a)-F(b)). Forgetting to integrate first.