What are applications to rates?

Because integration reverses differentiation, integrating a rate of change recovers the total change. If dVdt is the rate at which a tank fills, then _t_1^t_2(dV)/(dt)\,dt is the volume added between t_1 and t_2.

TASMath MethodsSyllabus dot point

What is integration, how is it the reverse of differentiation, and how do we find areas?

Find antiderivatives and definite integrals and use them to calculate areas and solve rate problems.

Integration is antidifferentiation; the fundamental theorem of calculus links the definite integral to the antiderivative, letting you compute areas under curves and recover quantities from their rates of change.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

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What this dot point is asking

Integration is the reverse process of differentiation. If $F'(x) = f(x)$ , then $F(x)$ is an antiderivative of $f(x)$ , and we write $\displaystyle\int f(x)\,dx = F(x) + C$ . The constant $C$ appears because differentiating any constant gives zero, so antiderivatives are only determined up to a constant.

The definite integral and the fundamental theorem

A definite integral has limits and produces a number.

The definite integral $\displaystyle\int_a^b f(x)\,dx$ gives the signed area between the curve $y=f(x)$ and the $x$ -axis from $x=a$ to $x=b$ . Area above the axis counts as positive and area below counts as negative.

Applications to rates

Because integration reverses differentiation, integrating a rate of change recovers the total change. If $\dfrac{dV}{dt}$ is the rate at which a tank fills, then $\displaystyle\int_{t_1}^{t_2}\frac{dV}{dt}\,dt$ is the volume added between $t_1$ and $t_2$ .

Worked example: area under a curve

Find the area enclosed between the curve $y = 6x - x^2$ and the $x$ -axis.

First find where the curve meets the $x$ -axis: $6x - x^2 = 0$ , so $x(6 - x) = 0$ , giving $x = 0$ and $x = 6$ . Between these, the parabola opens downward and lies above the axis, so the area is

A = \int_0^6 (6x - x^2)\,dx.

Find the antiderivative:

\int (6x - x^2)\,dx = 3x^2 - \frac{x^3}{3} + C.

Apply the limits (fundamental theorem):

A = \left[3x^2 - \frac{x^3}{3}\right]_0^6 = \left(3(36) - \frac{216}{3}\right) - 0 = 108 - 72 = 36.

The enclosed area is $36$ square units.

Summary

To integrate, reverse each differentiation rule and add $+C$ for indefinite integrals. For definite integrals, find an antiderivative and evaluate $F(b)-F(a)$ . For area problems, locate where the curve meets the axis, split at crossings, and take absolute values so positive and negative regions do not cancel. To go from a rate to a total, integrate the rate over the relevant interval.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC2 marksEvaluate the definite integral from 1 to 4 of (x^2 + 1)/x dx.

Show worked answer →

First simplify the integrand by dividing each term by x:
(x^2 + 1)/x = x + 1/x.

Antidifferentiate: integral of (x + 1/x) dx = x^2/2 + ln|x|.

Evaluate from 1 to 4:
[16/2 + ln 4] - [1/2 + ln 1] = (8 + ln 4) - (1/2 + 0) = 15/2 + ln 4.
So the value is 7.5 + ln 4 (about 8.886). The marks are for simplifying before integrating and for correctly substituting both limits, remembering ln 1 = 0.

2023 TASC5 marksa) Evaluate the definite integral from 1 to 2 of 1/x^2 dx. b) Solve for k: the integral from 1 to k of 1/x^2 dx = 0.99. c) What do you expect would happen to the integral from 1 to k of 1/x^2 dx as k tends to infinity?

Show worked answer →

The antiderivative of 1/x^2 = x^(-2) is -x^(-1) = -1/x.

a) Evaluate from 1 to 2: [-1/x] from 1 to 2 = (-1/2) - (-1/1) = -1/2 + 1 = 1/2.

b) Integral from 1 to k = [-1/x] from 1 to k = (-1/k) - (-1) = 1 - 1/k. Set equal to 0.99:
1 - 1/k = 0.99, so 1/k = 0.01, giving k = 100.

c) As k tends to infinity, 1/k tends to 0, so the integral tends to 1 - 0 = 1. The area under 1/x^2 from 1 to infinity converges to a finite value of 1.

2024 TASC8 marksA popcorn maker produces popcorn at a rate of dP/dt = -15t^2 + 60t + 5 sin(2 pi t), where P is the number of pieces produced in t minutes. It begins at t = 0 and ends when dP/dt = 0 once more. a) Find dP/dt at t = 1. b) Find the time at which production ends. c) Find P(t). d) Hence find the total amount of popcorn produced.

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a) Substitute t = 1: dP/dt = -15(1) + 60(1) + 5 sin(2 pi) = -15 + 60 + 0 = 45 pieces per minute (sin(2 pi) = 0).

b) Production ends when dP/dt = 0 again. Solving -15t^2 + 60t + 5 sin(2 pi t) = 0 with technology (the sin term is small) gives t = 4 minutes, the next positive root after t = 0.

c) Integrate the rate: P(t) = integral of (-15t^2 + 60t + 5 sin(2 pi t)) dt
= -5t^3 + 30t^2 - (5/(2 pi)) cos(2 pi t) + C.
Since P(0) = 0: 0 = 0 + 0 - 5/(2 pi) + C, so C = 5/(2 pi). The cos terms cancel for integer t.

d) Total = P(4) - P(0) = integral from 0 to 4 of dP/dt = -5(64) + 30(16) + 0 = -320 + 480 = 160 pieces (the periodic sin term integrates to 0 over the whole minutes).