Definite integrals, areas and volumes · Calculus (JAMB bridge)

Definite integrals

A definite integral has limits. Integrate (no $+ c$ is needed), then work out the result at the top limit and take away its value at the bottom limit:

\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q7 (a)

Evaluate $\displaystyle\int_1^2 (2x^3 - 4x + 3)\,dx$ .

Integrate
$\int (2x^3 - 4x + 3)\,dx = \frac{x^4}{2} - 2x^2 + 3x$ .
Think first. Integrate each term.
Top limit
$\frac{16}{2} - 8 + 6 = 6$ .
Think first. Put in x = 2.
Bottom limit
$\frac12 - 2 + 3 = \frac32$ .
Think first. Put in x = 1.
Subtract
$6 - \frac32 = 4\frac12$ .
Think first. 6 − 3/2 = ?

The area under a curve

The area between a curve, the $x$ -axis and the lines $x = a$ and $x = b$ is the definite integral $\int_a^b y\,dx$ . Think of it as the total of many thin strips.

Area under a curveThe area from x = a to x = b is the integral of y from a to b

Try it

Area under a curveAdd more strips

8.8594 strips9integral

strips = 4

The 4 strips add up to 8.859. The exact value is [x³/3] from 0 to 3 = 9 − 0 = 9: work out F at the top limit, then take away F at the bottom limit. Add more strips: the total closes in on the integral.

Add strips and watch their total close in on the integral. Try “x² − 2x from 0 to 3”: the part below the axis counts as negative and cancels the part above.

When the question gives no limits, they are where the curve crosses the $x$ -axis: solve $y = 0$ first.

Worked example · JAMB 2001

JAMB 2001 · UME · Q39

Find the area bounded by the curve $y = 4 - x^2$ and the $x$ -axis.

The limits
$x^2 = 4$ , so the curve crosses the axis at $x = -2$ and $x = 2$ .
Think first. Where does 4 − x² = 0?
Integrate
$\left[4x - \frac{x^3}{3}\right]_{-2}^{2}$ .
Think first. Integrate 4 − x².
Substitute
$\left(8 - \frac83\right) - \left(-8 + \frac83\right) = 16 - \frac{16}{3} = \frac{32}{3} = 10\frac23$ : option B.
Think first. F(2) − F(−2).

The area between a curve and a line

Find where they meet (these are the limits), then integrate top minus bottom.

Area between two graphsIntegrate (top − bottom) between the crossing points

Worked example · JAMB 2017

JAMB 2017 · UTME · Q33

Find the area bounded by the curves $y = 4 - x^2$ and $y = 2x + 1$ .

Where they meet
$x^2 + 2x - 3 = 0$ , so $(x + 3)(x - 1) = 0$ : $x = -3$ and $x = 1$ .
Think first. Solve 4 − x² = 2x + 1.
Top minus bottom
The curve is on top (at $x = 0$ : $4 > 1$ ). So integrate $(4 - x^2) - (2x + 1) = 3 - 2x - x^2$ .
Think first. Which is on top between −3 and 1?
Integrate
$\left(3 - 1 - \frac13\right) - (-9 - 9 + 9) = \frac53 + 9 = 10\frac23$ : option B.
Think first. [3x − x² − x³/3] from −3 to 1.

Volumes of revolution

Spin the region under $y = f(x)$ once round the $x$ -axis. Each thin slice is a disc of radius $y$ , so

V = \pi\int_a^b y^2\,dx

Volume of revolutionEach slice is a disc of radius y and area πy²

Distance from velocity

Velocity is the derivative of distance, so distance is the integral of velocity: the distance travelled from $t = a$ to $t = b$ is $\int_a^b v\,dt$ .

Your turn

JAMB 1995 · UME · Q41

Calculus (JAMB bridge)

Find the area bounded by the curve $y = 3x^2 - 2x + 1$ , the ordinates $x = 1$ and $x = 3$ , and the $x$ -axis.

Worked solution (try it first)

The area is $\int_1^3 (3x^2 - 2x + 1)\,dx = \left[x^3 - x^2 + x\right]_1^3$ .
At $x = 3$ : $27 - 9 + 3 = 21$ .
At $x = 1$ : $1 - 1 + 1 = 1$ .
Subtract: $21 - 1 = 20$ , option D.

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