True or False? When you subtract one function from the other to find the area enclosed between a curve and a line, it does not matter which way round you subtract, because an area cannot be negative.

False. Subtracting the wrong way round gives the negative of the area, and it will not be corrected for you. Always take the lower graph away from the upper one over the interval concerned, which means deciding from a sketch which of the two is on top.

What changes when both boundaries of a region are curves?

Nothing about the method: it is still the integral of upper minus lower , taken between the intersections. What is lost is the shortcut of using a triangle or trapezium formula for one boundary, since neither of them is a straight line any more.

Two curves meet at three points. How many integrals does the enclosed area need?

Two , one for each enclosed region: from the first intersection to the second, and from the second to the third. Each region is bounded separately, so each needs its own integral with its own pair of limits.

Why must you check which curve is on top for each region separately?

Because the two curves swap over at every point where they cross. A curve that was above before an intersection is below after it, so "upper minus lower" means a different subtraction in each region.

Why is a sketch essential here?

Because it is the only reliable way to see how many separate regions there are, and which curve is on top in each of them. The algebra gives you the intersections; only the picture tells you what to do with them.

True or False? The total area between two curves is the integral of their difference across the whole interval.

False. Wherever the curves swap over, the difference changes sign, so one region subtracts from another and the total comes out too small. Integrate each region separately, take each as a positive area, and then add them.

A LevelMathsCambridge (CIE)Pure 1FlashcardsIntegrationFurther Integration

Further Integration (Cambridge (CIE) A Level Maths: Pure 1): Flashcards

Exam code: 9709

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FrontArea Under a Curve
Define the area under a curve $y = f (x)$ between $x = a$ and $x = b$ .
BackArea Under a Curve
The area of the region bounded by four things:
the curve $y = f (x)$
the $x$ -axis
the vertical line $x = a$
the vertical line $x = b$
The $x$ -axis is the boundary that is easiest to forget: "under the curve" means between the curve and the $x$ -axis, not simply below the curve.
BackArea Under a Curve
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Cards in this collection (17)

Define the area under a curve $y = f (x)$ between $x = a$ and $x = b$ .
The area of the region bounded by four things:
- the curve $y = f (x)$
- the $x$ -axis
- the vertical line $x = a$
- the vertical line $x = b$
The $x$ -axis is the boundary that is easiest to forget: "under the curve" means between the curve and the $x$ -axis, not simply below the curve.
What has to be true about a curve between $x = a$ and $x = b$ for the area under it to be given by $\int_{a}^{b} f (x) d x$ ?
The curve must lie on or above the $x$ -axis across the whole of that interval.
Where it dips below, the integral counts that part as negative, so the value it returns is no longer the area.
A question asks for the area between a curve and the $x$ -axis but gives no limits. Where do the limits come from?
From the curve's $x$ -axis intercepts, which are the natural edges of the region.
Set $y = 0$ and solve. For example, $y = x (5 - x)$ gives $x = 0$ and $x = 5$ , so the area is $\int_{0}^{5} (5 x - x^{2}) d x$ .
True or False?
If a definite integral gives a negative value, you have made a mistake.
False.
A region lying below the $x$ -axis gives a negative integral. That is the integral behaving correctly, not an error.
An area cannot be negative, so take the modulus: an integral of $- \frac{22}{3}$ , for example, means an area of $\frac{22}{3}$ square units.
Before integrating to find the total area between a curve and the $x$ -axis, what must you check?
Whether the curve crosses the $x$ -axis anywhere inside the interval.
Solve $f (x) = 0$ to find out; each crossing splits the region into a separate piece, and each piece needs its own integral.
For example, $y = x^{3} - 12 x^{2} + 35 x$ crosses at $x = 0$ , $x = 5$ and $x = 7$ .
A region between a curve and the $x$ -axis lies partly above and partly below the axis. Why does a single definite integral across the whole interval not give its total area?
The part below the axis contributes a negative value, so it cancels part of the positive contribution from the part above. The result is smaller than the true area, and can even be zero.
Integrate each piece separately, take the modulus of each, then add; for example, two pieces giving $\frac{375}{4}$ and $- 8$ have a total area of $\frac{375}{4} + 8 = \frac{407}{4}$ square units.
Why do you need to solve the equations of a curve and a line simultaneously before finding the area enclosed between them?
Their points of intersection are the limits of integration: the enclosed region only exists between them.
Set the two expressions for $y$ equal and solve. For example, for $y = 5 + 4 x - x^{2}$ and $y = 10 - 2 x$ :
$10 - 2 x = 5 + 4 x - x^{2} \Rightarrow x^{2} - 6 x + 5 = 0$
so the limits are $x = 1$ and $x = 5$ .
The region enclosed between a curve and a line runs from $x = a$ to $x = b$ , with the curve above the line. How do you find its area by working out two separate areas?
Find the area under the curve and the area under the line over the same interval, then subtract the smaller from the larger:
$R = \int_{a}^{b} f (x) d x - \int_{a}^{b} g (x) d x$
Whichever graph is on top has the larger area underneath it, so with the curve on top it is the area under the line that is subtracted.
True or False?
To find the area under a straight line between two $x$ -values, you have to use definite integration.
False.
The region under a straight line is a rectangle, triangle or trapezium, so a basic area formula is usually quicker and less error-prone.
For example, the area under $y = 10 - 2 x$ from $x = 1$ to $x = 5$ is a triangle of base 4 and height 8, giving $\frac{1}{2} \times 4 \times 8 = 16$ .
How can you find the area enclosed between a curve and a line using only one integral?
Subtract the two functions before integrating, taking the lower graph away from the upper one, then integrate the result between the intersections.
For example, with the curve $y = 5 + 4 x - x^{2}$ above the line $y = 10 - 2 x$ between $x = 1$ and $x = 5$ :
$R = \int_{1}^{5} [(5 + 4 x - x^{2}) - (10 - 2 x)] d x = \int_{1}^{5} (- x^{2} + 6 x - 5) d x$
True or False?
When you subtract one function from the other to find the area enclosed between a curve and a line, it does not matter which way round you subtract, because an area cannot be negative.
False.
Subtracting the wrong way round gives the negative of the area, and it will not be corrected for you.
Always take the lower graph away from the upper one over the interval concerned, which means deciding from a sketch which of the two is on top.
Why must the whole of a line's equation be put in brackets when it is subtracted from a curve's equation?
Because every term of the line is being subtracted, not just the first one. For example:
$(5 + 4 x - x^{2}) - (10 - 2 x) = - x^{2} + 6 x - 5$
Without the brackets the $- 2 x$ would keep its own sign, giving $- x^{2} + 2 x - 5$ and an answer that is wrong from that line onwards.
What changes when both boundaries of a region are curves?
Nothing about the method: it is still the integral of upper minus lower, taken between the intersections.
What is lost is the shortcut of using a triangle or trapezium formula for one boundary, since neither of them is a straight line any more.
Two curves meet at three points. How many integrals does the enclosed area need?
Two, one for each enclosed region: from the first intersection to the second, and from the second to the third.
Each region is bounded separately, so each needs its own integral with its own pair of limits.
Why must you check which curve is on top for each region separately?
Because the two curves swap over at every point where they cross.
A curve that was above before an intersection is below after it, so "upper minus lower" means a different subtraction in each region.
Why is a sketch essential here?
Because it is the only reliable way to see how many separate regions there are, and which curve is on top in each of them.
The algebra gives you the intersections; only the picture tells you what to do with them.
True or False?
The total area between two curves is the integral of their difference across the whole interval.
False.
Wherever the curves swap over, the difference changes sign, so one region subtracts from another and the total comes out too small.
Integrate each region separately, take each as a positive area, and then add them.

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