Mathematics 9709 · Pure Mathematics 1

Definite integrals

A definite integral has two numbers on the sign (the limits). It gives you one single number, not a whole family of curves.
It is the tool behind every area and volume later in the chapter.

Integrate. Write the answer in square brackets with the limits. Then do top minus bottom: $F(b)-F(a)$ . You do not need $+c$ , because it cancels.

Integrate, then top minus bottom

The method is three steps, every time:

Integrate the function as usual, but write the answer inside square brackets and copy the limits onto them. No $+c$ — it cancels.
Substitute the top limit $b$ for every $x$ , to get $F(b)$ .
Substitute the bottom limit $a$ , and subtract: the answer is $F(b)-F(a)$ .

$\int_a^b \mathrm{f}(x)\,dx=\big[\,F(x)\,\big]_a^b=F(b)-F(a)$

Here $F$ is any function whose gradient is $\mathrm{f}$ . It is the indefinite integral, just without the constant.
The number you get is the area under the curve between the two limits. The picture below shows this:

A definite integral returns one number: the area under the curve between the two limits. Here the area under y = x² from 1 to 2 works out to exactly 7/3.

Why there is no +c

The constant cancels by itself, so you never write it.
Any constant you add goes into both $F(b)$ and $F(a)$ . When you subtract, it goes away.

See the +c cancel

Say you kept the constant, so your integral was $F(x)+c$ . Put in both limits and subtract:

$\big[F(b)+c\big]-\big[F(a)+c\big]=F(b)-F(a)+c-c$

The two copies of $c$ cancel, whatever $c$ was. You are left with $F(b)-F(a)$ . The $c$ never changes a definite answer, so there is no point writing it.

Two worked integrals

A fraction: split it into separate powers first, then integrate term by term.
A linear bracket: use the $(ax+b)^n$ rule from the last topic.

Worked example

Evaluate $\displaystyle\int_1^2 \dfrac{6x^4-1}{x^2}\,dx$

You cannot integrate a fraction term by term like this, so split it into separate powers of $x$ first:

$\dfrac{6x^4-1}{x^2}=6x^2-x^{-2}$

Now integrate, and write the result in square brackets with the limits:

$\int_1^2\big(6x^2-x^{-2}\big)\,dx=\Big[\,2x^3+x^{-1}\,\Big]_1^2$

Substitute the top limit, then the bottom limit, then subtract:

$=\Big(2(2)^3+\tfrac{1}{2}\Big)-\Big(2(1)^3+1\Big)$

$=\big(16+\tfrac12\big)-\big(2+1\big)=16\tfrac12-3$

Answer

13\tfrac{1}{2}

Worked example

Evaluate $\displaystyle\int_0^3 \sqrt{5x+1}\,dx$

The inside is the linear bracket $5x+1$ . So use the $(ax+b)^n$ rule: add one to the power, divide by the new power, then divide by $a=5$ :

$\int_0^3 (5x+1)^{\frac12}\,dx=\left[\dfrac{(5x+1)^{\frac32}}{5\left(\frac32\right)}\right]_0^3=\left[\tfrac{2}{15}(5x+1)^{\frac32}\right]_0^3$

At the top $x=3$ the bracket is $16$ ; at the bottom $x=0$ it is $1$ . So top minus bottom:

$\tfrac{2}{15}(16)^{\frac32}-\tfrac{2}{15}(1)^{\frac32}=\tfrac{2}{15}(64)-\tfrac{2}{15}=\tfrac{128}{15}-\tfrac{2}{15}=\tfrac{126}{15}$

Answer

8\tfrac{2}{5}

Where the marks go

Common mistake

It is top minus bottom,

F(b)-F(a)

, not the other way round. If you swap the limits, the whole answer changes sign. A right answer turns into a wrong one.

Common mistake

Keep the square brackets until you have integrated. Put in negative limits with care. Every

x

F(x)

becomes that negative number, with brackets around it. Dropping one minus sign is the common way to lose this mark.

Common mistake

Do not write

+c

on a definite integral. It always cancels. So it gets you nothing and just makes your working messy.

Now you try

Evaluate $\displaystyle\int_{-2}^{1}\dfrac{8}{(5-2x)^2}\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (rewrite as a negative power)

Rewrite as a power so you can use the linear-bracket rule: $8(5-2x)^{-2}$ . Integrate with new power $-1$ , dividing by $a=-2$ :

$\left[\dfrac{8(5-2x)^{-1}}{(-2)(-1)}\right]_{-2}^{1}=\left[\dfrac{4}{5-2x}\right]_{-2}^{1}$

Put the limits in carefully. Top $x=1$ : $\dfrac{4}{5-2(1)}=\dfrac{4}{3}$ . Bottom $x=-2$ : $\dfrac{4}{5-2(-2)}=\dfrac{4}{9}$ . Now top minus bottom:

$\dfrac{4}{3}-\dfrac{4}{9}=\dfrac{12}{9}-\dfrac{4}{9}$

Answer

\tfrac{8}{9}

PreviousIntegrating (ax + b)ⁿ NextArea under a curve

Definite integrals

A definite integral has two numbers on the sign (the limits). It gives you one single number, not a whole family of curves.
It is the tool behind every area and volume later in the chapter.

Integrate. Write the answer in square brackets with the limits. Then do top minus bottom: $F(b)-F(a)$ . You do not need $+c$ , because it cancels.

Integrate, then top minus bottom

The method is three steps, every time:

Integrate the function as usual, but write the answer inside square brackets and copy the limits onto them. No $+c$ — it cancels.
Substitute the top limit $b$ for every $x$ , to get $F(b)$ .
Substitute the bottom limit $a$ , and subtract: the answer is $F(b)-F(a)$ .

$\int_a^b \mathrm{f}(x)\,dx=\big[\,F(x)\,\big]_a^b=F(b)-F(a)$

Here $F$ is any function whose gradient is $\mathrm{f}$ . It is the indefinite integral, just without the constant.
The number you get is the area under the curve between the two limits. The picture below shows this:

A definite integral returns one number: the area under the curve between the two limits. Here the area under y = x² from 1 to 2 works out to exactly 7/3.

Why there is no +c

The constant cancels by itself, so you never write it.
Any constant you add goes into both $F(b)$ and $F(a)$ . When you subtract, it goes away.

See the +c cancel

Say you kept the constant, so your integral was $F(x)+c$ . Put in both limits and subtract:

$\big[F(b)+c\big]-\big[F(a)+c\big]=F(b)-F(a)+c-c$

The two copies of $c$ cancel, whatever $c$ was. You are left with $F(b)-F(a)$ . The $c$ never changes a definite answer, so there is no point writing it.

Two worked integrals

A fraction: split it into separate powers first, then integrate term by term.
A linear bracket: use the $(ax+b)^n$ rule from the last topic.

Worked example

Evaluate $\displaystyle\int_1^2 \dfrac{6x^4-1}{x^2}\,dx$

You cannot integrate a fraction term by term like this, so split it into separate powers of $x$ first:

$\dfrac{6x^4-1}{x^2}=6x^2-x^{-2}$

Now integrate, and write the result in square brackets with the limits:

$\int_1^2\big(6x^2-x^{-2}\big)\,dx=\Big[\,2x^3+x^{-1}\,\Big]_1^2$

Substitute the top limit, then the bottom limit, then subtract:

$=\Big(2(2)^3+\tfrac{1}{2}\Big)-\Big(2(1)^3+1\Big)$

$=\big(16+\tfrac12\big)-\big(2+1\big)=16\tfrac12-3$

Answer

13\tfrac{1}{2}

Worked example

Evaluate $\displaystyle\int_0^3 \sqrt{5x+1}\,dx$

The inside is the linear bracket $5x+1$ . So use the $(ax+b)^n$ rule: add one to the power, divide by the new power, then divide by $a=5$ :

$\int_0^3 (5x+1)^{\frac12}\,dx=\left[\dfrac{(5x+1)^{\frac32}}{5\left(\frac32\right)}\right]_0^3=\left[\tfrac{2}{15}(5x+1)^{\frac32}\right]_0^3$

At the top $x=3$ the bracket is $16$ ; at the bottom $x=0$ it is $1$ . So top minus bottom:

$\tfrac{2}{15}(16)^{\frac32}-\tfrac{2}{15}(1)^{\frac32}=\tfrac{2}{15}(64)-\tfrac{2}{15}=\tfrac{128}{15}-\tfrac{2}{15}=\tfrac{126}{15}$

Answer

8\tfrac{2}{5}

Where the marks go

Common mistake

It is top minus bottom,

F(b)-F(a)

, not the other way round. If you swap the limits, the whole answer changes sign. A right answer turns into a wrong one.

Common mistake

Keep the square brackets until you have integrated. Put in negative limits with care. Every

x

F(x)

becomes that negative number, with brackets around it. Dropping one minus sign is the common way to lose this mark.

Common mistake

Do not write

+c

on a definite integral. It always cancels. So it gets you nothing and just makes your working messy.

Now you try

Evaluate $\displaystyle\int_{-2}^{1}\dfrac{8}{(5-2x)^2}\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (rewrite as a negative power)

Rewrite as a power so you can use the linear-bracket rule: $8(5-2x)^{-2}$ . Integrate with new power $-1$ , dividing by $a=-2$ :

$\left[\dfrac{8(5-2x)^{-1}}{(-2)(-1)}\right]_{-2}^{1}=\left[\dfrac{4}{5-2x}\right]_{-2}^{1}$

Put the limits in carefully. Top $x=1$ : $\dfrac{4}{5-2(1)}=\dfrac{4}{3}$ . Bottom $x=-2$ : $\dfrac{4}{5-2(-2)}=\dfrac{4}{9}$ . Now top minus bottom:

$\dfrac{4}{3}-\dfrac{4}{9}=\dfrac{12}{9}-\dfrac{4}{9}$

Answer

\tfrac{8}{9}