Integrals

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Integrals are a central idea in calculus because they measure how quantities build up over an interval. Geometrically, a definite integral gives the signed area between a curve and the $x$ -axis. This lets students connect graphs, formulas, and real physical meaning. Integrals are used in physics, economics, biology, and engineering whenever change accumulates over time or space.

A definite integral adds up many tiny contributions, often written as slices of width $dx$ . If $f(x)$ is positive, the integral adds area above the axis, and if $f(x)$ is negative, it subtracts area below the axis. This is why integrals represent net change, not just total shaded region unless all values stay above the axis. The Fundamental Theorem of Calculus links this accumulation process to antiderivatives, making many integrals possible to compute exactly.

Key Facts

Definite integral: $\int_a^b f(x)\,dx$ gives the net signed area under $f(x)$ from $x = a$ to $x = b$ .
Accumulation function: $A(x) = \int_a^x f(t)\,dt$ measures how much $f$ has accumulated from $a$ to $x$ .
Fundamental Theorem of Calculus: if $F'(x) = f(x)$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ .
Net change formula: change in quantity = $\int_a^b \text{rate of change}\,dt$ .
If $f(x) > 0$ on $[a, b]$ , then $\int_a^b f(x)\,dx$ equals ordinary area under the curve.
Total area is found by splitting where $f(x)$ changes sign and adding absolute values: total area = $\int |f(x)|\,dx$ over the interval.

Vocabulary

Definite integral: A number that represents the accumulated net change or signed area of a function over an interval.
Antiderivative: A function $F(x)$ whose derivative is the original function $f(x)$ , so $F'(x) = f(x)$ .
Accumulation: The process of adding many small pieces together to find a total amount.
Net change: The overall change in a quantity after increases and decreases are both taken into account.
Signed area: Area counted as positive above the x-axis and negative below the x-axis.

Common Mistakes to Avoid

Treating every definite integral as ordinary area, which is wrong because parts below the $x$ -axis count as negative and reduce the result.
Forgetting the bounds of integration, which is wrong because an antiderivative alone does not give the accumulated value on a specific interval.
Mixing up total area with net change, which is wrong because total area requires splitting the interval and using absolute values where the function is negative.
Using the variable $x$ carelessly in an accumulation function, which is wrong because $A(x) = \int_a^x f(t)\,dt$ uses $t$ as a dummy variable and $x$ as the changing upper limit.

Practice Questions

1 Find $\int_0^3 (2x + 1)\,dx$ .
2 A particle has velocity $v(t) = 4t - 6$ meters per second for $0 \leq t \leq 5$ . Find the net change in position from $t = 0$ to $t = 5$ .
3 A graph of $f(x)$ lies above the $x$ -axis on $[1, 3]$ and below the $x$ -axis on $[3, 5]$ . Explain why $\int_1^5 f(x)\,dx$ can be smaller than the total shaded area between the curve and the axis.

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