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.

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Integration (AQA A Level Maths: Pure): Flashcards

Exam code: 7357

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FrontFundamental Theorem of Calculus
Define integration.

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Cards in this collection (32)

Define integration.
Integration is the inverse process of differentiation: it undoes differentiating.
So integrating $\frac{d y}{d x}$ takes you back to $y$ , and integrating $f' (x)$ takes you back to $f (x)$ .
This is what the Fundamental Theorem of Calculus states.
Why do $y = 5 x$ , $y = 5 x + 2$ and $y = 5 x - 3$ all have the same derivative?
Differentiating a constant term gives 0, because the graph of a constant is a horizontal line and a horizontal line has gradient 0.
The constant leaves no trace in the derivative, and all three give $\frac{d y}{d x} = 5$ .
Working backwards from $\frac{d y}{d x} = 5$ , the most you can say is $y = 5 x + c$ .
Define indefinite integration.
Integration carried out without limits, so the result is a function rather than a number.
Because the constant term cannot be recovered, the answer always ends with a constant of integration, $+ c$ , so that, for example:
$\int 2 x d x = x^{2} + c$
True or False?
Differentiating a function and then integrating the result always takes you back to exactly the function you started with.
False.
You get the original function back apart from its constant term, which the differentiation destroyed.
Differentiating $y = 6 x + 3$ gives $\frac{d y}{d x} = 6$ , and integrating $6$ gives $6 x + c$ ; nothing in $\frac{d y}{d x}$ records that the constant was $3$ .
This is why an indefinite integral always ends with $+ c$ .
Complete the notation for integrating $3 x^{2} + 5 x + 4$ with respect to $x$ , filling in the missing symbol at each end:
$______(3 x^{2} + 5 x + 4)______$
The completed integral is:
$\int (3 x^{2} + 5 x + 4) d x$
The integral sign and the $d x$ work as a pair. They say "integrate all of $(\dots)$ with respect to $x$ ", which is why a function of more than one term is written in brackets between them.
In the integral $\int (3 x^{2} + 5 x + 4) d x$ , what does the $d x$ tell you?
That you are integrating with respect to $x$ , so $x$ is the variable.
Any other letter in the function being integrated is treated as a constant, in the same way as an ordinary number.
Define the integrand.
The function that is being integrated: everything between the integral sign and the $d x$ .
In $\int (4 x^{3} - 7) d x$ the integrand is $4 x^{3} - 7$ .
Complete the rule for integrating a power of $x$ , filling in the missing index and the missing denominator:
$\int x^{n} d x = \frac{x^{______}}{______} + c$
The completed rule is:
$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$
In words: increase the power by 1, then divide by the new power.
When integrating, which single power of $x$ can the rule "increase the power by 1 and divide by the new power" not be used on, and why?
$x^{- 1}$ , which is $\frac{1}{x}$ .
Increasing $- 1$ by 1 gives a new power of 0, and the rule would then divide by 0, which is undefined.
$\int \frac{1}{x} d x$ does exist, but it is found a different way, later in the course.
True or False?
To integrate $3 x^{2} (2 x - 1)$ , you can integrate $3 x^{2}$ and $(2 x - 1)$ separately and then multiply the two answers together.
False.
There is no rule that lets you integrate a product one factor at a time. Expand the brackets first, then integrate term by term:
$\int 3 x^{2} (2 x - 1) d x = \int (6 x^{3} - 3 x^{2}) d x = \frac{3}{2} x^{4} - x^{3} + c$
Multiplying the separate integrals would give $x^{3} (x^{2} - x) + c$ , which is not the same.
Before you can integrate $\frac{{(x + 3)}^{2}}{\sqrt{x}}$ , what do you need to do to it?
Rewrite it as a sum of separate powers of $x$ , because that is the only form the power rule can be applied to.
Expand the numerator, write $\sqrt{x}$ as $x^{\frac{1}{2}}$ , then divide each term separately using the index laws:
$\frac{{(x + 3)}^{2}}{\sqrt{x}} = \frac{x^{2} + 6 x + 9}{x^{\frac{1}{2}}} = x^{\frac{3}{2}} + 6 x^{\frac{1}{2}} + 9 x^{- \frac{1}{2}}$
What is $\int 5 d x$ , and why?
$5 x + c$ .
A constant is a power of $x$ like any other, since $5 = 5 x^{0}$ . Increasing the power by 1 and dividing by the new power gives $\frac{5 x^{1}}{1} = 5 x$ .
After integrating a gradient function, how do you find the value of the constant of integration?
You need the coordinates of one point that the curve passes through.
Substitute them into the integrated equation and solve for $c$ . For example, if $y = \frac{3}{2} x^{4} - x^{3} + c$ passes through $(2, 6)$ :
$6 = \frac{3}{2} {(2)}^{4} - {(2)}^{3} + c \Rightarrow c = - 10$
When finding the constant of integration, why must you integrate the gradient function before you substitute the coordinates of a point on the curve?
The coordinates satisfy the equation of the curve, not the equation of its gradient function, so there is nothing useful to substitute them into until you have integrated.
The constant $c$ also only appears once you have integrated, so before that there is nothing to find.
Complete the definite integration form of the Fundamental Theorem of Calculus:
$\int_{a}^{b} f' (x) d x =______-______$
The completed statement is:
$\int_{a}^{b} f' (x) d x = f (b) - f (a)$
Integrate, then substitute the upper limit and subtract the value at the lower limit.
Define the limits of a definite integral.
The two values written on the integral sign, which fix the interval the integration is carried out over.
- the one at the bottom, $a$ in $\int_{a}^{b}$ , is the lower limit
- the one at the top, $b$ , is the upper limit
Why is there no " $+ c$ " in the answer to a definite integral?
Because it cancels out.
The constant appears at both limits, and the two copies are subtracted:
$(f (b) + c) - (f (a) + c) = f (b) - f (a)$
True or False?
To evaluate $\int_{1}^{3} 4 x^{3} d x$ , you substitute 3 and 1 into $4 x^{3}$ and subtract.
False.
You must integrate first, and substitute the limits into the integrated function:
$\int_{1}^{3} 4 x^{3} d x = {[x^{4}]}_{1}^{3} = 3^{4} - 1^{4} = 80$
Substituting into $4 x^{3}$ itself would give $108 - 4 = 104$ , which is not the value of the integral.
Your calculator evaluates a definite integral as $418.666667$ . Why might that not be an acceptable answer?
Because it is a rounded decimal, and many questions ask for an exact value.
Here the exact value is $\frac{1256}{3}$ , which only comes out of integrating by hand and substituting the limits.
A calculator value is still worth having as a check on your working.
What do the square brackets mean in ${[\frac{3}{4} x^{4} - 3 x^{2}]}_{2}^{4}$ ?
That the function has been integrated but the limits have not been substituted yet.
The limits are written outside the closing bracket, the lower one at the bottom and the upper one at the top. The next line substitutes them:
$(\frac{3}{4} {(4)}^{4} - 3 {(4)}^{2}) - (\frac{3}{4} {(2)}^{4} - 3 {(2)}^{2})$
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.

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