Integration · Calculus (JAMB bridge)

Reversing differentiation

Integration undoes differentiation. If $\frac{dy}{dx} = 2x$ , then $y$ could be $x^2$ , or $x^2 + 5$ , or $x^2 - 3$ : all of them have gradient $2x$ . So an integral always carries a constant, $+ c$ .

For powers of $x$ , reverse the power rule: add 1 to the power, then divide by the new power.

\int ax^n\,dx = \frac{ax^{n+1}}{n + 1} + c \qquad (n \ne -1)

∫ a xⁿ dx = a x^{n + 1} ⁄ (n + 1) + c

add 1 to the power, then divide by the new power; never forget + c (n ≠ −1)

Integrating a powerAdd 1 to the power, divide by the new power, add c

Negative powers work the same way:

\begin{aligned} \int \frac{3}{x^2}\,dx &= \int 3x^{-2}\,dx \\ &= \frac{3x^{-1}}{-1} + c = -\frac3x + c \end{aligned}

The one power the rule cannot do is $x^{-1}$ , because it would divide by 0. Instead, $\int \frac1x\,dx = \ln x + c$ .

Sine, cosine and (ax + b)

Reverse the derivatives of sine and cosine. When the inside is $ax + b$ , divide by $a$ . This reverses the chain rule, which multiplies by $a$ :

\begin{aligned} \int \cos ax\,dx &= \frac1a\sin ax + c \\ \int \sin ax\,dx &= -\frac1a\cos ax + c \\ \int (ax + b)^n\,dx &= \frac{(ax + b)^{n+1}}{a(n + 1)} + c \end{aligned}

Worked example · JAMB 2001

JAMB 2001 · UME · Q38

Evaluate $\displaystyle\int 2(2x - 3)^{\frac23}\,dx$ .

The pattern
- Use ${\int (ax + b)^n\,dx = \frac{(ax + b)^{n+1}}{a(n + 1)}}$ .
- Here ${a = 2}$ and ${n = \frac23}$ , so ${n + 1 = \frac53}$ .
Think first. Here a = 2 and n = 2/3.
Integrate
$\int 2(2x - 3)^{\frac23}\,dx = 2 \times \frac{(2x - 3)^{\frac53}}{2 \times \frac53}$ .
Think first. Divide by a(n + 1) = 2 × 5/3.
Simplify
$\frac35(2x - 3)^{\frac53} + k$ : option D.
Think first. 2 ÷ (10/3) = ?

Finding the constant

An integral gives a whole family of curves, one for each value of $c$ . If you also know one point on the curve, put it in to find $c$ .

The + c familyEvery curve y = F(x) + c has the same gradient; the given point picks out one

Try it

Finding the constant cSlide c until the curve passes through the point

2x + 1dy/dxx² + x + 0ynothrough (1, 4)?

c = 0

Integrating dy/dx = 2x + 1 gives y = x² + x + c. Every value of c gives a curve with the same gradients, just moved up or down, so the faint curves all fit. To pick the right one, put the point (1, 4) into y = x² + x + c and solve for c.

Slide $c$ and watch the curve move up and down without changing shape. Only one value of $c$ puts it through the point.

Worked example · JAMB 1995

JAMB 1995 · UME · Q40

Two variables $x$ and $y$ are such that $\frac{dy}{dx} = 4x - 3$ and $y = 5$ when $x = 2$ . Find $y$ in terms of $x$ .

Integrate
$y = 2x^2 - 3x + c$ .
Think first. Integrate 4x − 3.
Use the point
${5 = 8 - 6 + c}$ , so ${c = 3}$ .
Think first. Put x = 2, y = 5.
Answer
$y = 2x^2 - 3x + 3$ : option B.
Think first. Write y in terms of x.

Your turn

NECO 2023 · Paper 2 · Q8 (b)

(b)
Find the equation of the curve which passes through $(-2, 5)$ and has gradient $6x^2 + 8x - 3$ at any point.

Worked solution (try it first)

(b)

The gradient is $\frac{dy}{dx}$ , so integrate it: $y = \frac{6x^3}{3} + \frac{8x^2}{2} - 3x + c$
$= 2x^3 + 4x^2 - 3x + c$ .
The curve passes through $(-2, 5)$ : $2(-8) + 4(4) - 3(-2) + c = 5$ , so $-16 + 16 + 6 + c = 5$ and $c = -1$ .
The curve is $y = 2x^3 + 4x^2 - 3x - 1$ .

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