What is properties of the definite integral?

A few algebraic properties make definite integrals easier to handle and are regularly tested. Reversing the limits changes the sign, _a^b f = -_b^a f; an integral over a zero-width interval is zero, _a^a f = 0; a constant factor pulls out, _a^b kf = k_a^b f; a sum splits, _a^b(f + g) = _a^b f + _a^b g; and an interval can be split at any interior point, _a^b f = _a^c f + _c^b f. The last property is exactly what justifies splitting at axis-crossings or curve-crossings.

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.

TASC TCE 2023-style practice: Calculator-free. Evaluate _1^4 x^2 + 1x\,dx.

First simplify the integrand by dividing each term by x: x^2 + 1x = x + 1x. Antidifferentiate: (x + (1)/(x))dx = x^22 + ln|x|. Evaluate from 1 to 4: $[(16)/(2) + ln 4] - [(1)/(2) + ln 1] = (8 + ln 4) - 12 = (15)/(2) + ln 4.$ So the value is 7.5 + ln 4 ≈ 8.886. The marks are for simplifying before integrating and for substituting both limits, remembering ln 1 = 0.

TASC TCE 2023-style practice: Calculator-assumed. (a) Evaluate _1^2 1x^2\,dx. (b) Solve for k: _1^k 1x^2\,dx = 0.99. (c) What happens to _1^k 1x^2\,dx as k → ∞?

The antiderivative of 1x^2 = x^-2 is -x^-1 = -1x. (a) [-1x]_1^2 = -12 - (-1) = 12. (b) _1^k = [-1x]_1^k = 1 - 1k. Set equal to 0.99: 1 - 1k = 0.99, so 1k = 0.01, giving k = 100. (c) As k → ∞, 1k → 0, so the integral tends to 1. The area under 1x^2 from 1 to infinity converges to a finite value of 1.

§-Syllabus dot point

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 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Properties of the definite integral

A few algebraic properties make definite integrals easier to handle and are regularly tested. Reversing the limits changes the sign, $\displaystyle\int_{a}^{b} f = -\int_{b}^{a} f$ ; an integral over a zero-width interval is zero, $\displaystyle\int_{a}^{a} f = 0$ ; a constant factor pulls out, $\displaystyle\int_{a}^{b} kf = k\int_{a}^{b} f$ ; a sum splits, $\displaystyle\int_{a}^{b}(f + g) = \int_{a}^{b} f + \int_{a}^{b} g$ ; and an interval can be split at any interior point, $\displaystyle\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ . The last property is exactly what justifies splitting at axis-crossings or curve-crossings.

Finding a function from its derivative

A common application gives you a rate or a gradient function together with one known point, and asks for the original function. Integrate to introduce the constant $C$ , then use the point to solve for it. This "boundary condition" technique is the same one that fixes the constants in kinematics, and it turns an indefinite integral into a single definite answer.

Recovering a curve from its gradient

A curve has gradient $\dfrac{dy}{dx} = 3x^{2} - 4x$ and passes through the point $(2, 5)$ . Find $y$ as a function of $x$ .

Step 1: Integrate the gradient to find the family of curves

Antidifferentiate term by term. Because there are infinitely many antiderivatives differing by a constant, we include $+C$ to represent the whole family:

y = \int (3x^{2} - 4x)\,dx = x^{3} - 2x^{2} + C.

Step 2: Use the given point to find $C$

Substitute $x = 2$ and $y = 5$ into the expression above. This selects the single curve from the family that actually passes through $(2, 5)$ :

5 = (2)^{3} - 2(2)^{2} + C = 8 - 8 + C = C,

so $C = 5$ .

Final answer: $y = x^{3} - 2x^{2} + 5$ . The known point converts the family of antiderivatives into the one curve that passes through it.

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.

Step 1: Find where the curve meets the $x$ -axis

The limits of integration are the $x$ -values where the curve touches the axis, i.e. where $y = 0$ . Factoring reveals both roots:

6x - x^2 = 0 \quad\Rightarrow\quad x(6 - x) = 0 \quad\Rightarrow\quad x = 0 \text{ and } x = 6.

Between these values the downward-opening parabola lies entirely above the $x$ -axis, so the definite integral will give the correct positive area directly.

Step 2: Set up and find the antiderivative

Write the definite integral with the limits just found, then antidifferentiate term by term:

A = \int_0^6 (6x - x^2)\,dx, \qquad \int (6x - x^2)\,dx = 3x^2 - \frac{x^3}{3}.

Step 3: Apply the fundamental theorem

Evaluate the antiderivative at the upper limit and subtract its value at the lower limit:

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

Final answer: 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.

TCE 20232 marksCalculator-free. Evaluate

\displaystyle\int_{1}^{4} \dfrac{x^{2} + 1}{x}\,dx

Show worked answer →

First simplify the integrand by dividing each term by $x$ : $\dfrac{x^{2} + 1}{x} = x + \dfrac{1}{x}$ .

Antidifferentiate: $\displaystyle\int\left(x + \frac{1}{x}\right)dx = \frac{x^{2}}{2} + \ln|x|$ .

Evaluate from $1$ to $4$ :

\left[\frac{16}{2} + \ln 4\right] - \left[\frac{1}{2} + \ln 1\right] = (8 + \ln 4) - \tfrac{1}{2} = \frac{15}{2} + \ln 4.

So the value is

7.5 + \ln 4 \approx 8.886

. The marks are for simplifying before integrating and for substituting both limits, remembering

\ln 1 = 0

TCE 20235 marksCalculator-assumed. (a) Evaluate

\displaystyle\int_{1}^{2} \dfrac{1}{x^{2}}\,dx

. (b) Solve for

k

\displaystyle\int_{1}^{k} \dfrac{1}{x^{2}}\,dx = 0.99

. (c) What happens to

\displaystyle\int_{1}^{k} \dfrac{1}{x^{2}}\,dx

k \to \infty

Show worked answer →

The antiderivative of $\dfrac{1}{x^{2}} = x^{-2}$ is $-x^{-1} = -\dfrac{1}{x}$ .

(a) $\left[-\dfrac{1}{x}\right]_{1}^{2} = -\dfrac{1}{2} - (-1) = \dfrac{1}{2}$ .

(b) $\displaystyle\int_{1}^{k} = \left[-\dfrac{1}{x}\right]_{1}^{k} = 1 - \dfrac{1}{k}$ . Set equal to $0.99$ : $1 - \dfrac{1}{k} = 0.99$ , so $\dfrac{1}{k} = 0.01$ , giving $k = 100$ .

(c) As $k \to \infty$ , $\dfrac{1}{k} \to 0$ , so the integral tends to $1$ . The area under $\dfrac{1}{x^{2}}$ from $1$ to infinity converges to a finite value of $1$ .

TCE 20248 marksCalculator-assumed. A popcorn maker produces popcorn at a rate

\dfrac{dP}{dt} = -15t^{2} + 60t + 5\sin(2\pi t)

, where

P

is the number of pieces produced in

t

minutes, starting at

t = 0

and ending when

\dfrac{dP}{dt} = 0

once more. (a) Find

\dfrac{dP}{dt}

t = 1

. (b) Find the time production ends. (c) Find

P(t)

. (d) Hence find the total amount produced.

Show worked answer →

(a) Substitute $t = 1$ : $\dfrac{dP}{dt} = -15 + 60 + 5\sin(2\pi) = 45$ pieces per minute, since $\sin(2\pi) = 0$ .

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

(c) Integrate the rate: $P(t) = \displaystyle\int (-15t^{2} + 60t + 5\sin(2\pi t))\,dt = -5t^{3} + 30t^{2} - \frac{5}{2\pi}\cos(2\pi t) + C$ .
Since $P(0) = 0$ : $0 = -\dfrac{5}{2\pi} + C$ , so $C = \dfrac{5}{2\pi}$ .

(d) Total $= P(4) - P(0) = \displaystyle\int_{0}^{4}\frac{dP}{dt}\,dt = -5(64) + 30(16) + 0 = 160$ pieces (the periodic sine term integrates to $0$ over whole minutes).