Mathematics 9709 · Pure Mathematics 2

Definite integrals and the area under a curve

Every rule so far gives you an antiderivative.
A definite integral turns that into one number: integrate, then do top limit minus bottom limit.
That number has a picture. It is the area under the curve between $x=a$ and $x=b$ .

$\int_a^b f(x)\,dx=\big[F(x)\big]_a^b=F(b)-F(a)$ , where $F$ is any antiderivative. The $+c$ cancels when you subtract, so you never need it for a definite integral.

The integral is the area

Slide $a$ and $b$ below. The shaded patch is exactly what $\int_a^b y\,dx$ measures.
Pull the gap wider and the value grows. Close it up and it drops to zero.
The integral is just the area from a across to b.

lower aupper b

area ∫_a^b y dx = 18.17·from a = 1.50 to b = 6.00

Why F(b) − F(a) gives the area (the running-total idea)

Think of $F(x)$ as the running total of area collected from the start up to $x$ . Then the area from $a$ to $b$ is “total up to $b$ ” minus “total up to $a$ ”. That is $F(b)-F(a)$ . The part before $a$ is in both totals, so it cancels. This is the Fundamental Theorem of Calculus in one sentence. It is why anti-differentiating and finding area are the same thing.

Worked examples

Worked example

Area under $y = x^2 + e^{-2x}$ from $x = 1$ to $x = 2$ (textbook Worked example 5.3)

The shaded region between the curve, the x-axis and the lines $x=1$ , $x=2$ is just the definite integral of $y$ .

Area $=\displaystyle\int_1^2\!\big(x^2+e^{-2x}\big)\,dx$ . Integrate term by term: $x^2\to\tfrac{x^3}{3}$ , and $e^{-2x}\to-\tfrac12 e^{-2x}$ (divide by $a=-2$ ).
Antiderivative $\Big[\tfrac{x^3}{3}-\tfrac12 e^{-2x}\Big]_1^2$ .
Top limit $x=2$ : $\tfrac{8}{3}-\tfrac12 e^{-4}$ . Bottom $x=1$ : $\tfrac{1}{3}-\tfrac12 e^{-2}$ .
Subtract: $\tfrac{8}{3}-\tfrac13-\tfrac12 e^{-4}+\tfrac12 e^{-2}=\tfrac{7}{3}+\tfrac12 e^{-2}-\tfrac12 e^{-4}$ .

Answer

\tfrac{7}{3}+\tfrac12 e^{-2}-\tfrac12 e^{-4}

Worked example

Area between $y = 2 + \sec^2x$ , the axes and $x = \tfrac{\pi}{4}$

Area $=\displaystyle\int_0^{\pi/4}\!\big(2+\sec^2x\big)\,dx$ . Integrate: $2\to2x$ and $\sec^2x\to\tan x$ .
$\big[2x+\tan x\big]_0^{\pi/4}$ . Top: $\tfrac{\pi}{2}+\tan\tfrac{\pi}{4}=\tfrac{\pi}{2}+1$ ; bottom: $0+\tan0=0$ .

Answer

\tfrac{\pi}{2}+1

Where the marks go

Common mistake

Always do top minus bottom, not bottom minus top. Swap the order and the sign of your whole answer flips.

Common mistake

For an area, the curve should stay above the axis the whole way across

[a,b]

. If it dips below, that part of

\int_a^b y\,dx

comes out negative and partly cancels the rest. So split at the crossing point and add up the size of each piece. (In P3 the region is usually set up so this doesn't happen, but check.)

Now you try

Find the exact area of the region between $y=e^{2x}+\dfrac{1}{2x+1}$ , the $x$ -axis and the lines $x=0$ and $x=1$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Integrate each piece with its own $\tfrac1a$ : $\Big[\tfrac12 e^{2x}+\tfrac12\ln(2x+1)\Big]_0^1$ .

Top $x=1$ : $\tfrac12 e^{2}+\tfrac12\ln3$ ; bottom $x=0$ : $\tfrac12+\tfrac12\ln1=\tfrac12$ :

Answer

\tfrac12 e^{2}+\tfrac12\ln3-\tfrac12

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Definite integrals and the area under a curve

Every rule so far gives you an antiderivative.
A definite integral turns that into one number: integrate, then do top limit minus bottom limit.
That number has a picture. It is the area under the curve between $x=a$ and $x=b$ .

$\int_a^b f(x)\,dx=\big[F(x)\big]_a^b=F(b)-F(a)$ , where $F$ is any antiderivative. The $+c$ cancels when you subtract, so you never need it for a definite integral.

The integral is the area

Slide $a$ and $b$ below. The shaded patch is exactly what $\int_a^b y\,dx$ measures.
Pull the gap wider and the value grows. Close it up and it drops to zero.
The integral is just the area from a across to b.

lower aupper b

area ∫_a^b y dx = 18.17·from a = 1.50 to b = 6.00

Why F(b) − F(a) gives the area (the running-total idea)

Worked examples

Worked example

Area under $y = x^2 + e^{-2x}$ from $x = 1$ to $x = 2$ (textbook Worked example 5.3)

The shaded region between the curve, the x-axis and the lines $x=1$ , $x=2$ is just the definite integral of $y$ .

Area $=\displaystyle\int_1^2\!\big(x^2+e^{-2x}\big)\,dx$ . Integrate term by term: $x^2\to\tfrac{x^3}{3}$ , and $e^{-2x}\to-\tfrac12 e^{-2x}$ (divide by $a=-2$ ).
Antiderivative $\Big[\tfrac{x^3}{3}-\tfrac12 e^{-2x}\Big]_1^2$ .
Top limit $x=2$ : $\tfrac{8}{3}-\tfrac12 e^{-4}$ . Bottom $x=1$ : $\tfrac{1}{3}-\tfrac12 e^{-2}$ .
Subtract: $\tfrac{8}{3}-\tfrac13-\tfrac12 e^{-4}+\tfrac12 e^{-2}=\tfrac{7}{3}+\tfrac12 e^{-2}-\tfrac12 e^{-4}$ .

Answer

\tfrac{7}{3}+\tfrac12 e^{-2}-\tfrac12 e^{-4}

Worked example

Area between $y = 2 + \sec^2x$ , the axes and $x = \tfrac{\pi}{4}$

Area $=\displaystyle\int_0^{\pi/4}\!\big(2+\sec^2x\big)\,dx$ . Integrate: $2\to2x$ and $\sec^2x\to\tan x$ .
$\big[2x+\tan x\big]_0^{\pi/4}$ . Top: $\tfrac{\pi}{2}+\tan\tfrac{\pi}{4}=\tfrac{\pi}{2}+1$ ; bottom: $0+\tan0=0$ .

Answer

\tfrac{\pi}{2}+1

Where the marks go

Common mistake

Always do top minus bottom, not bottom minus top. Swap the order and the sign of your whole answer flips.

Common mistake

For an area, the curve should stay above the axis the whole way across

[a,b]

. If it dips below, that part of

\int_a^b y\,dx

comes out negative and partly cancels the rest. So split at the crossing point and add up the size of each piece. (In P3 the region is usually set up so this doesn't happen, but check.)

Now you try

Find the exact area of the region between $y=e^{2x}+\dfrac{1}{2x+1}$ , the $x$ -axis and the lines $x=0$ and $x=1$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Integrate each piece with its own $\tfrac1a$ : $\Big[\tfrac12 e^{2x}+\tfrac12\ln(2x+1)\Big]_0^1$ .

Top $x=1$ : $\tfrac12 e^{2}+\tfrac12\ln3$ ; bottom $x=0$ : $\tfrac12+\tfrac12\ln1=\tfrac12$ :

Answer

\tfrac12 e^{2}+\tfrac12\ln3-\tfrac12