On Finding Areas by Integrating

Excerpts

On Finding Areas by Integrating
Here we have the clue as to what to do; the definite integral between the two limits is *the difference* between the integral worked out for the superior limit and the integral worked out for the lower limit.
On Finding Areas by Integrating
There are $18$ whole squares and four triangles, each of which has an area equal to $1\frac{1}{2}$ squares; or, in total, $24$ squares. Hence $24$ is the numerical value of the integral of $\dfrac{x}{3}\, dx$ between the lower limit of $x = 0$ and the higher limit of $x = 12$.
On Finding Areas by Integrating
*N.B.*---Notice that in dealing with definite integrals the constant $C$ always disappears by subtraction.
On Finding Areas by Integrating
That is all very well; but a little thought will show you that something more must be done.
On Finding Areas by Integrating
By “quadratic mean” is denoted the square root of the mean of the squares of all the values between the limits considered.
On Finding Areas by Integrating
Instead of a strip of area, we consider a small triangle $OAB$, the angle at $O$ being $d\theta$, and we find the sum of all the little triangles making up the required area.
On Finding Areas by Integrating
The secret of solving this problem is to conceive the area as being divided up into a lot of narrow strips, each of them being of the width $dx$. The smaller we take $dx$, the more of them there will be between $x_1$ and $x_2$. Now, the whole area is clearly equal to the sum of the areas of all such strips.
On Finding Areas by Integrating
If then we were to subtract the smaller area from the larger, we should have left as a remainder the area $PQNM$, which is what we want. Here we have the clue as to what to do; the definite integral between the two limits is *the difference* between the integral worked out for the superior limit and the integral worked out for the lower limit.
On Finding Areas by Integrating
All integration between limits requires the difference between two values to be thus found. Also note that, in making the subtraction the added constant $C$ has disappeared.
On Finding Areas by Integrating
Now reckon out the area beneath the curve *by counting the little squares* below the line, from $x = 0$ as far as $x = 12$ on the right. There are $18$ whole squares and four triangles, each of which has an area equal to $1\frac{1}{2}$ squares; or, in total, $24$ squares.
On Finding Areas by Integrating
Consider an elementary zone or annulus of the surface ([fig:59]Fig. 59), of breadth $dr$, situated at a distance $r$ from the centre. We may consider the entire surface as consisting of such narrow zones, and the whole area $A$ will simply be the integral of all such elementary zones from centre to margin, that is, integrated from $r = 0$ to $r = R$.
On Finding Areas by Integrating
To find the mean ordinate, we shall have to find the area of the piece $OMN$, and then divide it by the length of the base $ON$. But before we can find the area we must ascertain the length of the base, so as to know up to what limit we are to integrate.
On Finding Areas by Integrating
In certain branches of physics, particularly in the study of alternating electric currents, it is necessary to be able to calculate the *quadratic mean* of a variable quantity. By “quadratic mean” is denoted the square root of the mean of the squares of all the values between the limits considered. Other names for the quadratic mean of any quantity are its “virtual” value, or its “r.m.s.” (meaning root-mean-square) value. The French term is *valeur efficace*.

Equations

On Finding Areas by Integrating
dS = y · dx
The area of one narrow vertical strip under the curve is its height times its width dx.
On Finding Areas by Integrating
\int^{x=x_2}_{x=x_1} y\, dx = y_2 - y_1
A definite integral between limits equals the difference between the integrated value at the superior limit and at the inferior limit.
On Finding Areas by Integrating
\text{area~$S$} = b(x_2 - x_1) + \frac{a}{3}(x_2^3 - x_1^3)
The area under the curve y = b + ax^2 between x_1 and x_2 is b(x_2 - x_1) + (a/3)(x_2^3 - x_1^3).
On Finding Areas by Integrating
\text{mean~$y$} = \frac{1}{x_1} \int^{x=x_1}_{x=0} y · dx
The mean ordinate of a curve from x = 0 to x = x_1 is the area under the curve divided by the length of the base.
On Finding Areas by Integrating
dA = 2 \pi r\, dr
The area of a narrow circular zone at radius r with width dr is its circumference 2 pi r times dr.
On Finding Areas by Integrating
A &= \pi R^2
The area of a circle of radius R is pi times R squared, found by integrating the zones from the centre to the margin.
On Finding Areas by Integrating
\tfrac{1}{2} \int^{\theta=\theta_2}_{\theta=\theta_1} r^2\, d\theta
The area between a polar curve and two rays at angles theta_1 and theta_2 is one half the integral of r squared d theta.
On Finding Areas by Integrating
2\pi y\, dx
The area of a narrow belt of a surface of revolution is its circumference 2 pi y times its width dx.
On Finding Areas by Integrating
&= b(1-\epsilon^{-a})
The area under the die-away curve y = b e^(-x) from x = 0 to x = a is b(1 - e^(-a)).
On Finding Areas by Integrating
\text{area of $1$~strip} = dS = y · dx.
The area of a narrow vertical strip under the curve is its average height y times its width dx, written as the small area element dS.
On Finding Areas by Integrating
\text{total area~$S$} = \int dS = \int y\, dx.
The total area S is the sum (integral) of all the strip areas, so S equals the integral of y dx.
On Finding Areas by Integrating
\int^{x=x_2}_{x=x_1} y\, dx = y_2 - y_1,
The definite integral between limits equals the difference between the integrated values at the superior and inferior limits.
On Finding Areas by Integrating
\text{area~$S$} = b(x_2 - x_1) + \frac{a}{3}(x_2^3 - x_1^3).
The area under the curve y = b + a x^2 between x_1 and x_2 is b(x_2 - x_1) + (a/3)(x_2^3 - x_1^3).
On Finding Areas by Integrating
\text{mean~$y$} = \frac{1}{x_1} \int^{x=x_1}_{x=0} y · dx.
The mean ordinate of a curve from x = 0 to x = x_1 is the integral of y dx over that range divided by the length x_1.
On Finding Areas by Integrating
dA = 2 \pi r\, dr.
The area of a narrow circular zone of breadth dr at distance r from the centre is its length 2πr times its breadth.
On Finding Areas by Integrating
A &= \pi R^2.
The area of a circle of radius R equals pi times R squared.
On Finding Areas by Integrating
\pi(r_2^2 - r_1^2)
The area of a plane ring is the outer circle's area minus the inner circle's area, pi times (r_2 squared minus r_1 squared).
On Finding Areas by Integrating
pv^n = c
The adiabatic curve of a perfect gas: pressure times volume raised to the index n stays constant.
On Finding Areas by Integrating
y &= b\epsilon^{-x}.
The die-away curve is given by y equal to b times epsilon to the power minus x.
On Finding Areas by Integrating
&= b(1-\epsilon^{-a}).
The area under the die-away curve from x = 0 to x = a equals b times (1 minus epsilon to the minus a).
On Finding Areas by Integrating
\sqrt[2] {\frac{1}{l} \int^l_0 y^2\, dx}.
The quadratic mean of a function y over x from 0 to l is the square root of the mean of y squared over that range.
On Finding Areas by Integrating
\text{quadratic mean} = \frac{1}{\sqrt 3}\, al.
For y = a x from 0 to l, the quadratic mean is a l divided by the square root of 3.
On Finding Areas by Integrating
\dfrac{2}{\sqrt 3}=1.155
The form-factor, the ratio of quadratic to arithmetical mean for y = ax, equals 2 over the square root of 3, about 1.155.
On Finding Areas by Integrating
\text{quadratic mean} = \sqrt[2]{\dfrac{l^{2a}}{2a+1}}.
For y = x^a from 0 to l, the quadratic mean is the square root of l to the power 2a over (2a+1).
On Finding Areas by Integrating
\tfrac{1}{2} \int^{\theta=\theta_2}_{\theta=\theta_1} r^2\, d\theta.
The area between a polar curve and two radii at angles theta_1 and theta_2 is one half the integral of r squared d theta.
On Finding Areas by Integrating
r=a(1+\cos \theta)
The polar equation of Pascal's snail gives r as a times one plus the cosine of theta.
On Finding Areas by Integrating
&= \frac{a^2(3\pi+8)}{8}.
The area of the first quadrant of Pascal's snail equals a squared times (3 pi + 8) over 8.
On Finding Areas by Integrating
y^2 = r^2 - x^2.
For a sphere of radius r, a slice at abscissa x has the squared ordinate y squared equal to r squared minus x squared.

Problems

Exercise XVIII

Exercise XVIII, problem 10a, p. 225
Find the area of the portion of the curve $xy=a$ included between $x=1$ and $x = a$. Find the mean ordinate between these limits.
Printed answer:
- $a\log_\epsilon a$, $\dfrac{a}{a - 1} \log_\epsilon a$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem a*log(a)
Exercise XVIII, problem 10b, p. 225
Find the area of the portion of the curve $xy=a$ included between $x=1$ and $x = a$. Find the mean ordinate between these limits.
Printed answer:
- $a\log_\epsilon a$, $\dfrac{a}{a - 1} \log_\epsilon a$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem a*log(a)/(a - 1)
Exercise XVIII, problem 11a, p. 225
Show that the quadratic mean of the function $y=\sin x$, between the limits of $0$ and $\pi$ radians, is $\dfrac{\sqrt2}{2}$. Find also the arithmetical mean of the same function between the same limits; and show that the form-factor is $=1.11$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: no printed answer to check
Exercise XVIII, problem 11b, p. 225
Show that the quadratic mean of the function $y=\sin x$, between the limits of $0$ and $\pi$ radians, is $\dfrac{\sqrt2}{2}$. Find also the arithmetical mean of the same function between the same limits; and show that the form-factor is $=1.11$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: no printed answer to check
Exercise XVIII, problem 11c, p. 225
Show that the quadratic mean of the function $y=\sin x$, between the limits of $0$ and $\pi$ radians, is $\dfrac{\sqrt2}{2}$. Find also the arithmetical mean of the same function between the same limits; and show that the form-factor is $=1.11$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: no printed answer to check
Exercise XVIII, problem 12a, p. 225
Find the arithmetical and quadratic means of the function $x^2+3x+2$, from $x=0$ to $x=3$.
Printed answer:
- $\text{Arithmetical mean} = 9.5$; $\text{quadratic mean} = 10.85$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes Rational(19,2)
Exercise XVIII, problem 12b, p. 225
Find the arithmetical and quadratic means of the function $x^2+3x+2$, from $x=0$ to $x=3$.
Printed answer:
- $\text{Arithmetical mean} = 9.5$; $\text{quadratic mean} = 10.85$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes sqrt(Rational(1177,10))
Exercise XVIII, problem 13a, p. 225
Find the quadratic mean and the arithmetical mean of the function $y=A_1 \sin x + A_1 \sin 3x$.
Printed answer:
- $\text{Quadratic mean} = \dfrac{1}{\sqrt{2}} \sqrt{A_1^2 + A_3^2}$; $\text{arithmetical mean} = 0$. The first involves a somewhat difficult integral, and may be stated thus: By definition the quadratic mean will be 12 _0^2 (A_1 x + A_3 3x)^2 dx. %[** TN: Moved period out of radicand] Now the integration indicated by (A_1^2 ^2 x + 2A_1 A_3 x 3x + A_3^2 ^2 3x) dx is more readily obtained if for $\sin^2 x$ we write 1 - 2x2. For $2\sin x \sin 3x$ we write $\cos 2x - \cos 4x$; and, for $\sin^2 3x$, 1 - 6x2. Making these substitutions, and integrating, we get (see cosax) A_1^22 ( x - 2x2 ) + A_1 A_3 ( 2x2 - 4x4 ) + A_3^22 ( x - 6x6 ). At the lower limit the substitution of $0$ for $x$ causes all this to vanish, whilst at the upper limit the substitution of $2\pi$ for $x$ gives $A_1^2 \pi + A_3^2 \pi$. And hence the answer follows.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem sqrt(A_1**2 + A_3**2)/sqrt(2)
Exercise XVIII, problem 13b, p. 224
Find the quadratic mean and the arithmetical mean of the function $y=A_1 \sin x + A_1 \sin 3x$.
Printed answer:
- $\text{arithmetical mean} = 0$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 0
Exercise XVIII, problem 14a, p. 225
A certain curve has the equation $y=3.42\epsilon^{0.21x}$. Find the area included between the curve and the axis of $x$, from the ordinate at $x=2$ to the ordinate at $x = 8$. Find also the height of the mean ordinate of the curve between these points.
Printed answer:
- Area is $62.6$ square units. Mean ordinate is $10.42$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 3.42/0.21*(exp(1.68) - exp(0.42))
Exercise XVIII, problem 14b, p. 225
A certain curve has the equation $y=3.42\epsilon^{0.21x}$. Find the area included between the curve and the axis of $x$, from the ordinate at $x=2$ to the ordinate at $x = 8$. Find also the height of the mean ordinate of the curve between these points.
Printed answer:
- Area is $62.6$ square units. Mean ordinate is $10.42$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem 3.42/0.21*(exp(1.68) - exp(0.42))/6
Exercise XVIII, problem 15, p. 225
Show that the radius of a circle, the area of which is twice the area of a polar diagram, is equal to the quadratic mean of all the values of $r$ for that polar diagram.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
Exercise XVIII, problem 16, p. 225
Find the volume generated by the curve $y=±\dfrac{x}{6}\sqrt{x(10-x)}$ rotating about the axis of $x$.
Printed answer:
- $436.3$. (This solid is pear shaped.)
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 1250*pi/9
Exercise XVIII, problem 1a, p. 224
Find the area of the curve $y=x^2+x-5$ between $x=0$ and $x=6$, and the mean ordinates between these limits.
Printed answer:
- $\text{Area} = 60$; $\text{mean ordinate} = 10$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 60
Exercise XVIII, problem 1b, p. 224
Find the area of the curve $y=x^2+x-5$ between $x=0$ and $x=6$, and the mean ordinates between these limits.
Printed answer:
- $\text{Area} = 60$; $\text{mean ordinate} = 10$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 10
Exercise XVIII, problem 2, p. 224
Find the area of the parabola $y=2a\sqrt x$ between $x=0$ and $x=a$. Show that it is two-thirds of the rectangle of the limiting ordinate and of its abscissa.
Printed answer:
- $\text{Area} = \frac{2}{3}$ of $a × 2a \sqrt{a}$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem Rational(2,3)*(a*(2*a*sqrt(a)))
Exercise XVIII, problem 3a, p. 224
Find the area of the positive portion of a sine curve and the mean ordinate.
Printed answer:
- $\text{Area} = 2$; $\text{mean ordinate} = \dfrac{2}{\pi} = 0.637$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 2
Exercise XVIII, problem 3b, p. 224
Find the area of the positive portion of a sine curve and the mean ordinate.
Printed answer:
- $\text{Area} = 2$; $\text{mean ordinate} = \dfrac{2}{\pi} = 0.637$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 2/pi
Exercise XVIII, problem 4a, p. 224
Find the area of the positive portion of the curve $y=\sin^2 x$, and find the mean ordinate.
Printed answer:
- $\text{Area} = 1.57$; $\text{mean ordinate} = 0.5$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes pi/2
Exercise XVIII, problem 4b, p. 224
Find the area of the positive portion of the curve $y=\sin^2 x$, and find the mean ordinate.
Printed answer:
- $\text{Area} = 1.57$; $\text{mean ordinate} = 0.5$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 1/2
Exercise XVIII, problem 5a, p. 224
Find the area included between the two branches of the curve $y=x^2 ± x^{\efrac{5}{2}}$ from $x=0$ to $x=1$, also the area of the positive portion of the lower branch of the curve (see [fig:30]Fig. 30, fig:30). %[ ** Page, xref]
Printed answer:
- $0.572$, $0.0476$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: PASS-LOOSE Rational(4,7)
Exercise XVIII, problem 5b, p. 224
Find the area included between the two branches of the curve $y=x^2 ± x^{\efrac{5}{2}}$ from $x=0$ to $x=1$, also the area of the positive portion of the lower branch of the curve (see [fig:30]Fig. 30, fig:30). %[ ** Page, xref]
Printed answer:
- $0.572$, $0.0476$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes Rational(1,21)
Exercise XVIII, problem 6, p. 224
Find the volume of a cone of radius of base $r$, and of height $h$.
Printed answer:
- $\text{Volume} = \pi r^2 \dfrac{h}{3}$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem pi*r**2*h/3
Exercise XVIII, problem 7, p. 224
Find the area of the curve $y=x^3-\log_\epsilon x$ between $x=0$ and $x=1$.
Printed answer:
- $1.25$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes Rational(5,4)
Exercise XVIII, problem 8, p. 224
Find the volume generated by the curve $y=\sqrt{1+x^2}$, as it revolves about the axis of $x$, between $x=0$ and $x=4$.
Printed answer:
- $79.4$.
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem 76*pi/3
Exercise XVIII, problem 9a, p. 225
Find the volume generated by a sine curve revolving about the axis of $x$. Find also the area of its surface.
Printed answer:
- $\text{Volume} = 4.9348$; $\text{area of surface} = 12.57$ (from $0$ to $\pi$).
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes pi**2/2
Exercise XVIII, problem 9b, p. 225
Find the volume generated by a sine curve revolving about the axis of $x$. Find also the area of its surface.
Printed answer:
- $\text{Volume} = 4.9348$; $\text{area of surface} = 12.57$ (from $0$ to $\pi$).
unverified: no computed check settled this one (yet)
How it was checked
- evaluate: the printed answer does not match the problem 2*pi*(sqrt(2) + asinh(1))