What should you check before answering an area question?

Whether the question wants the value of an integral or an area . Wherever part of the region lies below the axis, those are two different numbers.

Which function must come first inside the integral?

The one that is above the other across the interval. Putting them the other way round gives the negative of the area.

How do you find the area between two curves that cross at three points?

Split at the middle intersection and integrate each region separately. In each one subtract the lower function from the upper, remembering that the two swap over at the crossing.

Why does the absolute value method need no knowledge of the zeros?

Because taking the modulus already turns every negative piece positive, wherever the zeros happen to fall. Splitting the interval does the same job by hand, but it needs the zeros found first.

What is the first thing to do in an arc length question?

Differentiate the equation of the curve to get the derivative the formula needs. Then substitute that derivative and the two endpoint coordinates into the integral.

True or False? The arc length of a curve between two points equals the straight-line distance between them.

False. Arc length is measured along the curve, so it is at least the straight-line distance and usually more. The two agree only when the curve is itself a straight line.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 8: Applications of IntegrationAreas & Arc Lengths

Areas & Arc Lengths (College Board AP® Calculus BC): Flashcards

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FrontArea Between a Curve & x-Axis
Fill in the missing condition for the area between $y = f (x)$ and the $x$ -axis:
$\int_{a}^{b} f (x) d x$ equals that area as long as $f (x)______0$ across the whole interval

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Cards in this collection (36)

Fill in the missing condition for the area between $y = f (x)$ and the $x$ -axis:
$\int_{a}^{b} f (x) d x$ equals that area as long as $f (x)______0$ across the whole interval
The completed condition is $f (x) \geq 0$ across the whole interval.
Wherever that holds, the definite integral and the area are the same number.
Why does a definite integral give an area?
Because $f (x) \cdot Δ x$ is the area of a thin rectangle of height $f (x)$ and width $Δ x$ .
The integral sums those rectangles as their width shrinks towards zero.
True or False?
A region lying below the $x$ -axis has a negative area.
False.
The integral comes out negative there, but an area cannot be negative.
Take the absolute value of the integral to get the area.
What do you do when the limits of integration are not given?
They are usually the $x$ -axis intercepts, so set $y = 0$ and solve.
Remember that the $y$ -axis, that is $x = 0$ , can also be one of the boundaries.
Find the area between $y = 9 - x^{2}$ and the $x$ -axis from $x = 1$ to $x = 2$ .
It is $\frac{20}{3}$ square units.
Evaluating ${[9 x - \frac{1}{3} x^{3}]}_{1}^{2}$ gives $\frac{46}{3} - \frac{26}{3}$ .
What should you check before answering an area question?
Whether the question wants the value of an integral or an area.
Wherever part of the region lies below the axis, those are two different numbers.
Find the area enclosed by $y = x^{2} - 4 x + 3$ and the $x$ -axis.
It is $\frac{4}{3}$ square units.
The curve meets the axis at $x = 1$ and $x = 3$ , and $\int_{1}^{3} (x^{2} - 4 x + 3) d x = - \frac{4}{3}$ , so take the absolute value.
What are the limits for an area against the $y$ -axis?
They are $y$ -coordinates, and the integration is with respect to $y$ .
If they are not given, set $x = 0$ and solve to find the $y$ -axis intercepts.
Fill in the two missing steps for an area against the $y$ -axis, given $y = 2 + \sqrt{x + 4}$ :
first rearrange into the form $x =______$ and then integrate with respect to $______$ between the two limits
The completed method is: first rearrange into the form $x = {(y - 2)}^{2} - 4$ and then integrate with respect to $y$ between the two limits.
Isolating the root and squaring both sides is what turns the equation round.
True or False?
An area against the $y$ -axis needs the equation rearranged into $x$ in terms of $y$ .
True.
The integration is with respect to $y$ , so the integrand must be a function of $y$ .
Only an equation already given in the form $x = g (y)$ can be integrated straight away.
When does the integral equal the area rather than its negative?
When the region lies to the right of the $y$ -axis, so that $g (y) \geq 0$ throughout.
A region to the left has $x < 0$ and gives a negative integral, whose absolute value is the area.
Find the area enclosed by $y = 2 + \sqrt{x + 4}$ , the $y$ -axis and $y = 6$ .
It is $\frac{32}{3}$ square units.
Rearranging gives $x = {(y - 2)}^{2} - 4$ , and the integral runs from the curve's $y$ -intercept at 4 up to 6.
Why must the differential match the variable in the integrand?
Because $d y$ says the integration is with respect to $y$ , so the integrand has to be a function of $y$ .
Writing $\int g (y) d x$ mixes two variables and means nothing.
Find the area enclosed by $x = y^{2} - 7 y + 10$ and the $y$ -axis.
It is $\frac{9}{2}$ square units.
Setting $x = 0$ gives limits $y = 2$ and $y = 5$ , and the integral comes to $- \frac{9}{2}$ because the region lies to the left of the axis.
Fill in the two missing functions in the area between two curves, where $f$ is the upper curve:
$\int_{a}^{b} (______-______) d x$ for curves $y = f (x)$ and $y = g (x)$
The completed integral is $\int_{a}^{b} (f (x) - g (x)) d x$ .
It is the difference of the two separate area integrals, gathered into one.
Which function must come first inside the integral?
The one that is above the other across the interval.
Putting them the other way round gives the negative of the area.
True or False?
You must worry about parts of the region falling below the $x$ -axis.
False.
If $f$ is above $g$ then $f (x) - g (x) \geq 0$ everywhere on the interval, whatever the two curves are doing relative to the axis.
The integral gives the correct area either way.
How do you find the limits when the region is enclosed by two curves?
Solve $f (x) = g (x)$ to find the $x$ -values where they intersect.
Those intersections are the limits of integration.
Find the area enclosed by $y = 2 x^{2} - 4 x + 2$ and $y = - 2 x^{2} + 8 x - 6$ .
It is $\frac{2}{3}$ square units.
They meet at $x = 1$ and $x = 2$ , and the downward parabola is on top, so $\int_{1}^{2} (- 4 x^{2} + 12 x - 8) d x$ gives the answer.
How does the rule change when the curves are functions of $y$ ?
Integrate $\int_{a}^{b} (f (y) - g (y)) d y$ between $y$ -limits.
The function that must come first is the one further from the $y$ -axis, which replaces 'above' as the test.
How do you find the limits for an area between two curves in terms of $y$ ?
Solve $f (y) = g (y)$ to find the $y$ -values where they intersect.
Those $y$ -values are the limits, exactly as the $x$ -values are in the other orientation.
The curves $x = \ln (8 y)$ and $x = \ln (\frac{2}{y})$ bound a region from $y = \frac{1}{2}$ to $y = 2$ . Set up the area integral.
It is $\int_{\frac{1}{2}}^{2} (\ln (8 y) - \ln (\frac{2}{y})) d y$ .
The first curve is further from the $y$ -axis so it comes first, and the logarithm laws simplify the integrand to $2 \ln (2 y)$ .
When does an area need more than one definite integral?
When the region is partly above and partly below the $x$ -axis, or when two curves cross more than twice.
A single integral across the whole interval would let the parts cancel each other out.
Fill in the two missing words in the method for a region partly above and partly below the axis:
split the interval at the $______$ of the function, then add the integrals of the parts above to the $______$ value of the integrals of the parts below
The completed method is: split the interval at the zeros of the function, then add the integrals of the parts above to the absolute value of the integrals of the parts below.
Working out the areas above and below the axis separately is what stops them cancelling.
Find the total area enclosed by $y = x (x - 1) (x + 2)$ and the $x$ -axis.
It is $\frac{37}{12}$ square units.
The zeros are $- 2$ , $0$ and $1$ ; the integral is $\frac{8}{3}$ on the first piece and $- \frac{5}{12}$ on the second, so the total is $\frac{8}{3} + \frac{5}{12}$ .
True or False?
With $\int_{a}^{b} | f (x) - g (x) | d x$ you must still put the upper curve first.
False.
The absolute value makes the order irrelevant, since $| f (x) - g (x) | = | g (x) - f (x) |$ .
That is the convenience of the method: you need not work out which curve is on top where.
How do you find the area between two curves that cross at three points?
Split at the middle intersection and integrate each region separately.
In each one subtract the lower function from the upper, remembering that the two swap over at the crossing.
Fill in the missing function in the single-integral method for a total area:
$Area = \int_{a}^{b}______d x$ for a region between $y = f (x)$ and the $x$ -axis
The completed integral is $\int_{a}^{b} | f (x) | d x$ .
The absolute value turns every negative part of $f$ positive, so the pieces cannot cancel.
Find the total area enclosed by $y = x^{3} - 12 x^{2} + 35 x$ and the $x$ -axis.
It is $101.75$ square units.
The zeros are $0$ , $5$ and $7$ , and $\int_{0}^{7} | x^{3} - 12 x^{2} + 35 x | d x = \frac{407}{4}$ .
Why does the absolute value method need no knowledge of the zeros?
Because taking the modulus already turns every negative piece positive, wherever the zeros happen to fall.
Splitting the interval does the same job by hand, but it needs the zeros found first.
Fill in the two missing parts of the arc length of $y = f (x)$ from $x = a$ to $x = b$ :
$L = \int_{a}^{b} \sqrt{______+ {(______)}^{2}} d x$ for a smooth planar curve
The completed formula is $L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ .
The limits are the $x$ -coordinates of the two endpoints of the arc.
What is the first thing to do in an arc length question?
Differentiate the equation of the curve to get the derivative the formula needs.
Then substitute that derivative and the two endpoint coordinates into the integral.
True or False?
The arc length of a curve between two points equals the straight-line distance between them.
False.
Arc length is measured along the curve, so it is at least the straight-line distance and usually more.
The two agree only when the curve is itself a straight line.
What is the arc length formula for a curve given as $x = f (y)$ ?
It is $L = \int_{c}^{d} \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$ .
Here $c$ and $d$ are the $y$ -coordinates of the endpoints, and the derivative is $\frac{d x}{d y}$ .
Write the arc length of $y = 1 + \ln x$ from $x = 1$ to $x = 10$ as an integral.
It is $\int_{1}^{10} \sqrt{\frac{x^{2} + 1}{x^{2}}} d x$ .
Here $\frac{d y}{d x} = \frac{1}{x}$ , and $1 + \frac{1}{x^{2}}$ combines over the common denominator $x^{2}$ .
Find the length of $y = \frac{2}{3} x^{\frac{3}{2}}$ from $x = 0$ to $x = 8$ .
It is $\frac{52}{3}$ units.
Since $\frac{d y}{d x} = x^{\frac{1}{2}}$ the integrand simplifies to $\sqrt{1 + x}$ , and the substitution $u = 1 + x$ gives $\frac{2}{3} (27 - 1)$ .

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