Integration – OCR FSMQ Additional Maths

1 Reversing Differentiation

CA9: be aware that integration is the reverse of differentiation. The question integration answers is: "what function would I have to differentiate to get this?"

CA8 — the integration rule

$\displaystyle\int kx^n \,\mathrm{d}x = \dfrac{kx^{n+1}}{n+1} + c$
Raise the power by one, then divide by the new power.

Differentiating multiplies by the old power and drops it by one. Integrating raises the power by one and divides by the new one. Each step of one is precisely undone by the other.

Integrand	Integral	Check by differentiating
$x^3$	$\dfrac{x^4}{4} + c$	$\dfrac{4x^3}{4} = x^3$ ✓
$6x^2$	$2x^3 + c$	$6x^2$ ✓
$x$	$\dfrac{x^2}{2} + c$	$x$ ✓
$5$	$5x + c$	$5$ ✓
$-2x^4$	$-\dfrac{2x^5}{5} + c$	$-2x^4$ ✓

A constant integrates to $kx$. Think of $5$ as $5x^0$: raising the power gives $5x^1$, and dividing by $1$ leaves $5x$.

Every indefinite integral needs $+c$. Differentiation destroys constant terms, so integration cannot know what the constant was. Omitting $+c$ costs a mark every time.

2 Indefinite Integrals (CA10)

Indefinite versus definite

Indefinite: $\displaystyle\int \mathrm{f}(x)\,\mathrm{d}x$ — no limits, answer is a function, needs $+c$
Definite: $\displaystyle\int_a^b \mathrm{f}(x)\,\mathrm{d}x$ — with limits, answer is a number, no $c$ needed

Worked Example 1 — A polynomial

Find $\displaystyle\int \left(6x^2 - 4x + 3\right)\mathrm{d}x$.

①$6x^2 \to \dfrac{6x^3}{3} = 2x^3$

②$-4x \to -\dfrac{4x^2}{2} = -2x^2$

③$3 \to 3x$

④$= 2x^3 - 2x^2 + 3x + c$

⑤Check by differentiating: $6x^2 - 4x + 3$ ✓

Always check by differentiating. It takes five seconds and catches every arithmetic slip.

Worked Example 2 — Expand first

Find $\displaystyle\int x(x-3)^2\,\mathrm{d}x$.

①There is no product rule for integration either — expand:

②$(x-3)^2 = x^2 - 6x + 9$, so $x(x-3)^2 = x^3 - 6x^2 + 9x$

③$\displaystyle\int \left(x^3 - 6x^2 + 9x\right)\mathrm{d}x = \dfrac{x^4}{4} - 2x^3 + \dfrac{9x^2}{2} + c$

3 Finding a Curve From Its Gradient Function

The specification's note under CA8 says you should "be able to find the equation of a curve, given its gradient function and one point". The extra point is what determines $c$.

Integrate the gradient function, including $+c$.
Substitute the coordinates of the known point.
Solve for $c$.
Write out the full equation with $c$ replaced by its value.

Worked Example 3 — Finding the curve

A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x$ and passes through $(2, 5)$. Find its equation.

①$y = \displaystyle\int \left(3x^2 - 4x\right)\mathrm{d}x = x^3 - 2x^2 + c$

②Substitute $(2,5)$: $5 = 8 - 8 + c$

③$c = 5$

④$y = x^3 - 2x^2 + 5$

⑤Check: differentiating gives $3x^2 - 4x$ ✓ and $y(2) = 8 - 8 + 5 = 5$ ✓

Geometrically, the gradient function describes a whole family of parallel curves, one for each value of $c$. The point selects the single member that passes through it.

Worked Example 4 — Through the origin

A curve has $\mathrm{f}'(x) = 4x - 1$ and passes through the origin. Find $\mathrm{f}(x)$.

①$\mathrm{f}(x) = 2x^2 - x + c$

②Through the origin means $\mathrm{f}(0) = 0$: $0 = 0 - 0 + c$, so $c = 0$.

③$\mathrm{f}(x) = 2x^2 - x$

Passing through the origin is the one case where $c = 0$ — but you should still show the substitution.

4 Definite Integrals (CA11)

Evaluating a definite integral

$\displaystyle\int_a^b \mathrm{f}(x)\,\mathrm{d}x = \Big[\mathrm{F}(x)\Big]_a^b = \mathrm{F}(b) - \mathrm{F}(a)$
where $\mathrm{F}$ is any integral of $\mathrm{f}$ — "top minus bottom"

Why $c$ disappears. Including it would give $\left(\mathrm{F}(b)+c\right) - \left(\mathrm{F}(a)+c\right)$, and the two $c$'s cancel. So there is no need to write it.

Worked Example 5 — A definite integral in full

Evaluate $\displaystyle\int_1^3 \left(2x + 1\right)\mathrm{d}x$.

①Integrate: $\Big[x^2 + x\Big]_1^3$

②Top limit: $3^2 + 3 = 12$

③Bottom limit: $1^2 + 1 = 2$

④$12 - 2 = 10$

Show the square brackets and both substitutions. The specification is explicit: writing the answer straight from a calculator's integral function "would therefore not be awarded full marks" when detailed reasoning is required.

Worked Example 6 — The specification's own example

Evaluate $\displaystyle\int_0^1 \left(4x^3 + x - 1\right)\mathrm{d}x$.

①$\Bigg[x^4 + \dfrac{x^2}{2} - x\Bigg]_0^1$

②At $x = 1$: $1 + \dfrac12 - 1 = \dfrac12$

③At $x = 0$: $0$

④$= \dfrac{1}{2}$

The specification's exemplar for "detailed reasoning" is exactly this kind of question: the bracket line and the substitution line are the evidence of method, and both are needed.

Worked Example 7 — Negative limits

Evaluate $\displaystyle\int_{-2}^{1} 3x^2 \,\mathrm{d}x$.

①$\Big[x^3\Big]_{-2}^{1}$

②At $x = 1$: $1$

③At $x = -2$: $(-2)^3 = -8$

④$1 - (-8) = 9$

Subtracting a negative adds. This is the commonest arithmetic error in definite integration. Write out the bracket: $1 - (-8)$, not $1 - 8$.

Two useful properties

$\displaystyle\int_a^b = -\int_b^a$ (swapping the limits changes the sign)
$\displaystyle\int_a^a = 0$ (equal limits give zero)

Worked Example 8 — A negative answer

Evaluate $\displaystyle\int_0^2 \left(x^2 - 4\right)\mathrm{d}x$ and interpret the result.

①$\Bigg[\dfrac{x^3}{3} - 4x\Bigg]_0^2$

②At $x=2$: $\dfrac83 - 8 = -\dfrac{16}{3}$

③At $x=0$: $0$

④$= -\dfrac{16}{3}$

⑤The answer is negative because the curve $y = x^2 - 4$ lies below the $x$-axis throughout $0 < x < 2$.

The area is $\tfrac{16}{3}$ square units; the integral is $-\tfrac{16}{3}$. Keeping that distinction clear matters for area questions.

5 Quick Reference

The rule

$kx^n \to \dfrac{kx^{n+1}}{n+1} + c$.

Constants

$k \to kx$.

$+c$

Compulsory for every indefinite integral.

Brackets

Expand before integrating.

Check

Differentiate your answer.

Finding $c$

Substitute the given point.

Definite

$\mathrm{F}(b) - \mathrm{F}(a)$, no $c$.

Show the brackets

Method marks live there.

Negative limits

Subtracting a negative adds.

Negative answer

Means the curve is below the axis.

6 Practice Questions

Question 1

Find $\displaystyle\int x^5 \,\mathrm{d}x$.

▶ Show solution

$\dfrac{x^6}{6} + c$

Question 2

Find $\displaystyle\int \left(8x^3 - 6x + 5\right)\mathrm{d}x$.

▶ Show solution

$\dfrac{8x^4}{4} - \dfrac{6x^2}{2} + 5x + c$

$= 2x^4 - 3x^2 + 5x + c$

Question 3

Evaluate $\displaystyle\int_0^3 2x \,\mathrm{d}x$.

▶ Show solution

$\Big[x^2\Big]_0^3 = 9 - 0 = 9$

Question 4

Evaluate $\displaystyle\int_1^2 \left(3x^2 + 2\right)\mathrm{d}x$.

▶ Show solution

$\Big[x^3 + 2x\Big]_1^2$

$= (8 + 4) - (1 + 2) = 12 - 3 = 9$

Question 5

Find $\displaystyle\int (2x+1)^2 \,\mathrm{d}x$.

▶ Show solution

Expand: $(2x+1)^2 = 4x^2 + 4x + 1$

$\displaystyle\int \left(4x^2+4x+1\right)\mathrm{d}x = \dfrac{4x^3}{3} + 2x^2 + x + c$

Question 6

A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - 2$ and passes through $(1, 4)$. Find its equation.

▶ Show solution

$y = 3x^2 - 2x + c$

$4 = 3 - 2 + c$, so $c = 3$.

$y = 3x^2 - 2x + 3$

Question 7

Evaluate $\displaystyle\int_{-1}^{2} \left(x^2 + 1\right)\mathrm{d}x$.

▶ Show solution

$\Bigg[\dfrac{x^3}{3} + x\Bigg]_{-1}^{2}$

At $x=2$: $\dfrac83 + 2 = \dfrac{14}{3}$

At $x=-1$: $-\dfrac13 - 1 = -\dfrac43$

$\dfrac{14}{3} - \left(-\dfrac43\right) = \dfrac{18}{3} = 6$

Question 8

A curve has $\mathrm{f}'(x) = 3x^2 + 4x - 1$ and $\mathrm{f}(0) = 7$. Find $\mathrm{f}(2)$.

▶ Show solution

$\mathrm{f}(x) = x^3 + 2x^2 - x + c$

$\mathrm{f}(0) = c = 7$, so $\mathrm{f}(x) = x^3 + 2x^2 - x + 7$.

$\mathrm{f}(2) = 8 + 8 - 2 + 7 = 21$

Question 9

Find the value of $k$ for which $\displaystyle\int_0^k 4x \,\mathrm{d}x = 50$.

▶ Show solution

$\Big[2x^2\Big]_0^k = 2k^2$

$2k^2 = 50$, so $k^2 = 25$ and $k = \pm 5$.

Taking the positive value (so that $k$ is above the lower limit $0$): $k = 5$.

Note $k = -5$ also satisfies the algebra, since $\int_0^{-5} 4x\,\mathrm{d}x = 2(25) = 50$ as well — the sign is lost in the squaring. Most questions intend $k > 0$.

Question 10

A curve $C$ has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x + 9$ and passes through the point $(0, 4)$.

(a) Find the equation of $C$. (b) Find the coordinates of the stationary points of $C$ and determine their nature. (c) Evaluate $\displaystyle\int_0^4 \dfrac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x$ and explain what the answer represents. (d) Verify your answer to (c) using the equation from (a).

▶ Show solution

(a) $y = \displaystyle\int \left(3x^2 - 12x + 9\right)\mathrm{d}x = x^3 - 6x^2 + 9x + c$

Substituting $(0,4)$: $4 = 0 - 0 + 0 + c$, so $c = 4$.

$y = x^3 - 6x^2 + 9x + 4$

(b) $3x^2 - 12x + 9 = 3\left(x^2-4x+3\right) = 3(x-1)(x-3) = 0$

$x = 1$ or $x = 3$.

At $x=1$: $y = 1 - 6 + 9 + 4 = 8$. At $x=3$: $y = 27 - 54 + 27 + 4 = 4$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 12$.

At $x=1$: $-6 < 0$, so $(1, 8)$ is a maximum.

At $x=3$: $6 > 0$, so $(3, 4)$ is a minimum.

(c) $\displaystyle\int_0^4 \left(3x^2 - 12x + 9\right)\mathrm{d}x = \Big[x^3 - 6x^2 + 9x\Big]_0^4$

At $x=4$: $64 - 96 + 36 = 4$. At $x=0$: $0$.

The integral equals $4$.

Integrating a gradient function between two limits gives the total change in $y$ over that interval — not an area under $C$ itself. So the value $4$ means $y$ has risen by $4$ between $x=0$ and $x=4$.

(d) From (a), $y(0) = 4$ and $y(4) = 64 - 96 + 36 + 4 = 8$.

Change in $y = 8 - 4 = 4$ ✓ — exactly the value of the integral.

This is the fundamental relationship between the two halves of calculus: differentiating $y$ gives the rate of change, and integrating that rate back over an interval recovers the net change in $y$. Notice that the constant $4$ played no part in (c), which is precisely why an indefinite integral cannot determine it.

Integration (CA8–CA11) · OCR FSMQ Additional Maths · Created with MathJax