1.08PureStretch

Definite Integrals

Q: Why is there no +,c in a definite integral?

It would be added in F(b) and subtracted in F(a), so it always cancels.

Q: Can a definite integral be negative?

Yes. Where the curve is below the x-axis the integral counts area as negative. The next lesson handles this for areas.

An indefinite integral is a family of functions. A $\textbf{definite integral}$ , written with limits like $\int_{1}^{3}$ , is a single number: evaluate the integral at the top limit, subtract its value at the bottom limit, and the unknown constant disappears. That number turns out to be a signed area — which is what makes integration so useful.

40 min Integration

The big picture

The Fundamental Theorem of Calculus says that if $\mathrm{F}$ is any antiderivative of $\mathrm{f}$ , then $\int_{a}^{b} \mathrm{f}(x)\,\mathrm{d}x = \mathrm{F}(b) - \mathrm{F}(a)$ , and this equals the limit of the sum of thin rectangle areas under the curve. So an area problem — adding infinitely many infinitely thin strips — becomes an algebra problem. This lesson focuses on evaluating definite integrals accurately: the notation, the properties that let you split and reverse limits, exact answers involving $\ln$ and $\mathrm{e}$ , and finding unknown limits or constants. The next lesson applies all of it to areas.

What you'll be able to do

Evaluate definite integrals using $\big[\mathrm{F}(x)\big]_{a}^{b} = \mathrm{F}(b) - \mathrm{F}(a)$
Understand the definite integral as the limit of a sum
Use properties of definite integrals, including reversing and splitting limits
Give exact answers and solve for unknown limits or constants

Evaluating between limits

Integrate as usual (no $+\,c$ needed), write the result in square brackets with the limits attached, then substitute: top limit minus bottom limit.

The constant would appear in both $\mathrm{F}(b)$ and $\mathrm{F}(a)$ and cancel, which is why it is omitted.

\int_{a}^{b} \mathrm{f}(x)\,\mathrm{d}x = \big[\mathrm{F}(x)\big]_{a}^{b} = \mathrm{F}(b) - \mathrm{F}(a)

The Fundamental Theorem of Calculus.

Example 1Evaluate

\int_{1}^{4} \left(3\sqrt{x} - \dfrac{2}{x^{2}}\right)\mathrm{d}x

1Integrate:

\big[2x^{3/2} + 2x^{-1}\big]_{1}^{4}

2At

x = 4

2(8) + \tfrac24 = 16.5

3At

x = 1

2 + 2 = 4

Answer

16.5 - 4 = 12.5

Tip — Substitute into the whole bracket at each limit, and wrap each evaluation in brackets before subtracting. Sign errors from a negative lower evaluation are extremely common.

The integral as a limit of a sum

Divide the interval from $a$ to $b$ into $n$ strips of width $\delta x$ . A rectangle of height $\mathrm{f}(x)$ on each strip has area $\mathrm{f}(x)\,\delta x$ , and their total approximates the area under the curve.

As the strips become infinitely thin, the sum becomes exact. The integral sign $\int$ is an elongated S, for "sum", and $\mathrm{d}x$ is the limiting strip width.

\int_{a}^{b} \mathrm{f}(x)\,\mathrm{d}x = \lim_{\delta x \to 0} \sum_{x=a}^{b} \mathrm{f}(x)\,\delta x

Definite integral as a limit of a sum.

Why this works

This is why integration measures accumulated quantities of every kind — distance from speed, work from force, mass from density — not just area. Anything that is "rate times a small width, added up" is an integral.

Properties of definite integrals

Reversing the limits changes the sign: $\int_{b}^{a} \mathrm{f}(x)\,\mathrm{d}x = -\int_{a}^{b} \mathrm{f}(x)\,\mathrm{d}x$ .

Limits can be split: $\int_{a}^{c} \mathrm{f}\,\mathrm{d}x = \int_{a}^{b} \mathrm{f}\,\mathrm{d}x + \int_{b}^{c} \mathrm{f}\,\mathrm{d}x$ . This is needed when a function changes formula, or when a region crosses the $x$ -axis.

Constants factor out, and the integral of a sum is the sum of integrals.

Example 1Given

\int_{0}^{3} \mathrm{g}(x)\,\mathrm{d}x = 7

and

\int_{0}^{5} \mathrm{g}(x)\,\mathrm{d}x = 11

, find

\int_{5}^{3} 2\mathrm{g}(x)\,\mathrm{d}x

\int_{3}^{5} \mathrm{g}\,\mathrm{d}x = 11 - 7 = 4

2Reversing the limits:

\int_{5}^{3} \mathrm{g}\,\mathrm{d}x = -4

3Multiply by 2.

Answer

-8

Exact answers and unknowns

Integrals of $\tfrac1x$ and $\mathrm{e}^{kx}$ give exact answers in terms of $\ln$ and $\mathrm{e}$ . Simplify using log laws: $\ln 6 - \ln 2 = \ln 3$ .

If a limit or a constant in the integrand is unknown, evaluate the integral in terms of it and solve the resulting equation.

Example 1Find the exact value of

\int_{1}^{3} \left(\dfrac{4}{x} + \mathrm{e}^{2x}\right)\mathrm{d}x

1Integrate:

\big[4\ln x + \tfrac12\mathrm{e}^{2x}\big]_{1}^{3}

2Top:

4\ln 3 + \tfrac12\mathrm{e}^{6}

. Bottom:

0 + \tfrac12\mathrm{e}^{2}

Answer

4\ln 3 + \tfrac12(\mathrm{e}^{6} - \mathrm{e}^{2})

Example 2Given

\int_{1}^{k} (2x - 1)\,\mathrm{d}x = 12

with

k > 1

, find

k

\big[x^{2} - x\big]_{1}^{k} = (k^{2} - k) - 0 = 12

k^{2} - k - 12 = 0

, so

(k - 4)(k + 3) = 0

Answer

k = 4

(since

k > 1

)

Tip — "Exact" means no decimals: leave $\ln$ , $\mathrm{e}$ , $\pi$ and surds in the answer, simplified.

Think like an examiner

Show the square-bracket line with limits before substituting.
Bracket each evaluation before subtracting.
Simplify exact answers with log laws.
Reject unknown values that violate given conditions, and say why.

Common misconceptions

Subtracting top minus bottom the wrong way round.

\mathrm{F}(b) - \mathrm{F}(a)

: top limit first.

Assuming

\mathrm{F}(0) = 0

without checking.

Evaluate at the lower limit even when it is 0 —

\mathrm{e}^{0} = 1

and

\cos 0 = 1

Adding

+\,c

and leaving it in a definite integral.

The constant cancels; omit it.

Evaluating trig integrals with the calculator in degrees.

Limits involving

\pi

are radians.

Definite integrals

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

Evaluation.

\int_{b}^{a} \mathrm{f}\,\mathrm{d}x = -\int_{a}^{b} \mathrm{f}\,\mathrm{d}x

Reversing limits.

\int_{a}^{c} = \int_{a}^{b} + \int_{b}^{c}

Splitting.

Stretch yourself

Show that $\int_{0}^{\pi/3} \left(\sin 2x + \cos x\right)\mathrm{d}x = \dfrac{3 + 2\sqrt3}{4}$ .

Hint — Integrate each term, then use exact values of $\cos\tfrac{2\pi}{3}$ and $\sin\tfrac{\pi}{3}$ . Take care at the lower limit.

Reveal solution

SolutionIntegrate:

\big[-\tfrac12\cos 2x + \sin x\big]_{0}^{\pi/3}

. At the top:

-\tfrac12\cos\tfrac{2\pi}{3} + \sin\tfrac{\pi}{3} = -\tfrac12\left(-\tfrac12\right) + \tfrac{\sqrt3}{2} = \tfrac14 + \tfrac{\sqrt3}{2}

. At the bottom:

-\tfrac12\cos 0 + \sin 0 = -\tfrac12

. Subtracting:

\tfrac14 + \tfrac{\sqrt3}{2} + \tfrac12 = \tfrac34 + \tfrac{\sqrt3}{2} = \dfrac{3 + 2\sqrt3}{4}

, as required.

\blacksquare

Questions students ask

Why is there no

+\,c

in a definite integral?

It would be added in

\mathrm{F}(b)

and subtracted in

\mathrm{F}(a)

, so it always cancels.

Can a definite integral be negative?

Yes. Where the curve is below the

x

-axis the integral counts area as negative. The next lesson handles this for areas.

What does

\mathrm{d}x

mean in a definite integral?

It indicates the variable of integration and, conceptually, the limiting width of the thin strips being summed.

Do I need radians?

Yes, for any integral involving trigonometric functions.

Key takeaways

A definite integral gives a number: $\mathrm{F}(b) - \mathrm{F}(a)$ .
It is the limit of a sum of thin strips, which is why it measures area and accumulation.
Reversing limits changes the sign; limits can be split at any point.
Give exact answers with $\ln$ , $\mathrm{e}$ and surds simplified.
Unknown limits or constants lead to equations to solve.