Sec 4 · Additional Mathematics

Definite integrals and areas

Practise definite integrals and areas with step-by-step help. Free account.

1. Definite integrals

Rule	Formula
evaluate	$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$
swap the limits	$\int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx$
join the ranges	$\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx$

F(x) = x³ − x²: upper limit, then minus the lower

\displaystyle \int_{1}^{3} (3x^{2} - 2x) \, dx

e.g.

\displaystyle (3^{3} - 3^{2}) - (1^{3} - 1^{2})

\displaystyle = 18

Exact answers: keep e and π

\displaystyle \int_{0}^{1} e^{2x} \, dx

e.g.

\displaystyle \dfrac{1}{2}e^{2} - \dfrac{1}{2}e^{0}

\displaystyle = \dfrac{1}{2}(e^{2} - 1)

📖 Worked example

Evaluate $\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \cos x \, dx$ , giving your answer in exact form.

Step 1 of 2

Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits

\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \cos x \, dx = \Big[\sin x\Big]_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}

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2. Areas

Region	Area
above the $x$ -axis	$\int_{a}^{b} y \, dx$
below the $x$ -axis	$\left\|\int_{a}^{b} y \, dx\right\|$
between the curve and the $y$ -axis	$\int_{c}^{d} x \, dy$
between a curve and a line	$\int_{a}^{b} (y_{\text{top}} - y_{\text{bottom}}) \, dx$

Left: y = (x + 1)(x − 2) with the region below the x-axis between x = −1 and x = 2 shaded; its area is the size of the integral. Right: the region between the line y = x + 2 (on top) and the curve y = x² − 4x + 6 from x = 1 to x = 4 shaded.

Below the axis: the integral is negative

\displaystyle \int_{-1}^{2} (x + 1)(x - 2) \, dx = -\dfrac{9}{2}

area $\dfrac{9}{2}$ units²

Line minus curve, between the meeting points

\displaystyle \int_{1}^{4} (-x^{2} + 5x - 4) \, dx = \dfrac{9}{2}

$x + 2 - (x^{2} - 4x + 6)$ , which is $-x^{2} + 5x - 4$

📖 Worked example

Find the area of the region enclosed by the curve $y = x^{2} - 4x + 6$ and the line $y = x + 2$ .

Step 1 of 4

Where do they meet? Set the curve equal to the line, bring everything to one side and factorise

x^{2} - 4x + 6 = x + 2

x^{2} - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1

x = 4

Practise this →

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1. Definite integrals

Rule	Formula
evaluate	$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$
swap the limits	$\int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx$
join the ranges	$\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx$

F(x) = x³ − x²: upper limit, then minus the lower

\displaystyle \int_{1}^{3} (3x^{2} - 2x) \, dx

e.g.

\displaystyle (3^{3} - 3^{2}) - (1^{3} - 1^{2})

\displaystyle = 18

Exact answers: keep e and π

\displaystyle \int_{0}^{1} e^{2x} \, dx

e.g.

\displaystyle \dfrac{1}{2}e^{2} - \dfrac{1}{2}e^{0}

\displaystyle = \dfrac{1}{2}(e^{2} - 1)

📖 Worked example

Evaluate $\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \cos x \, dx$ , giving your answer in exact form.

Step 1 of 2

Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits

\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \cos x \, dx = \Big[\sin x\Big]_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}

Practise this →

2. Areas

Region	Area
above the $x$ -axis	$\int_{a}^{b} y \, dx$
below the $x$ -axis	$\left\|\int_{a}^{b} y \, dx\right\|$
between the curve and the $y$ -axis	$\int_{c}^{d} x \, dy$
between a curve and a line	$\int_{a}^{b} (y_{\text{top}} - y_{\text{bottom}}) \, dx$

Below the axis: the integral is negative

\displaystyle \int_{-1}^{2} (x + 1)(x - 2) \, dx = -\dfrac{9}{2}

area $\dfrac{9}{2}$ units²

Line minus curve, between the meeting points

\displaystyle \int_{1}^{4} (-x^{2} + 5x - 4) \, dx = \dfrac{9}{2}

$x + 2 - (x^{2} - 4x + 6)$ , which is $-x^{2} + 5x - 4$

📖 Worked example

Find the area of the region enclosed by the curve $y = x^{2} - 4x + 6$ and the line $y = x + 2$ .

Step 1 of 4

Where do they meet? Set the curve equal to the line, bring everything to one side and factorise

x^{2} - 4x + 6 = x + 2

x^{2} - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1

x = 4

Practise this →