List of Notes - Category: Applications of Integrals
Now, when we know what definite integral is, we can calculate area between curves.
Consider the region `S` that lies between two curves `y=f(x)` and `y=g(x)` and between the vertical lines `x=a` and `x=b`, where `f` and `g` are continuous functions and `f(x)>=g(x)` for all `x` in `[a,b]`.
The Evaluation Theorem says that if f is continuous on [a,b], then `int_a^bf(x)dx=F(b)-F(a)` where F is any antiderivative of f. This means that F'=f and equation can be rewritten as `int_a^bF'(x)dx=F(b)-F(a)` .
In trying to find volume of the solid we use same approach as with area problem.
We divide solid into `n` pieces, approximate volume of each piece, take sum of volumes and then take limit as `n->oo`.
Sometimes it is very hard to use Method of Disks/Rings to obtain volume of solid of revolution.
But that's not a problem. We can use any shape for the cross section as long as it can be expanded or contracted to completely cover the solid.
We know how to find the volume of a solid of revolution obtained by rotating a region about a horizontal or vertical line (see Method of Disks/Rings and Method of Cylindrical Shells). But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical?
If the curve is a polygon, we can easily find its length; we just add the lengths of the line segments that form the polygon. (We can use the distance formula to find the distance between the endpoints of each segment.)
It is easy to compute average value of finitely many values `y_1`, `y_2`,...,`y_n`: `y_(ave)=(y_1+y_2+...+y_n)/n`.
But how to compute average of infinitely many values? In general, let's try to compute the average value of a function `y=f(x)`, `a<=x<=b`.
Economists often use demand function `D(p)`. The demand function `D(p)` is the quantity of goods that can be sold when price of 1 unit is `p`. Usually, the bigger price, the lower demand, so the demand function is a decreasing function.
Work is the total amount of effort required to perform a task.
In general, if an object moves along a straight line with position function `s(t)`, then the force `F` on the object (in the same direction) is defined by Newton’s Second Law of Motion as the product of its mass and its acceleration: `F=m(d^2s)/(dt^2)`.
Suppose that a thin horizontal plate with area `A` square meters is submerged in a fluid of density `rho` kilograms per cubic meter at a depth `d` meters below the surface of the fluid.
The the volume of fluid above plate is `V=Ad`, so its mass is `m=rho V=rho Ad`.
In this note we will try to find a point P on which a thin plate of any given shape balances horizontally. This point is called the center of mass (or center of gravity) of the plate.
We first consider the simpler situation. Suppose that the rod lies along the x-axis with `m_1` at `x_1` and `m_2` at `x_2` and the center of mass at `barx`. According to the Law of Lever We see
Suppose we consider the cholesterol level of a person chosen at random from a certain age group, or the height of an adult female chosen at random, or the lifetime of a randomly chosen battery of a certain type. Such quantities are called continuous random variables because their values actually range over an interval of real numbers, although they might be measured or recorded only to the nearest integer. Random variable is usually denoted by X. We assume that probability that continuous variable will take ceratain value a equals 0, that is `P(X=a)=0`.
Consider the distance problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. If the velocity remains constant, then the distance problem is easy to solve by means of the formula `s=vt` .
We know that area under the curve `y=F(x)` is `A=int_a^b F(x)dx` where `f(x)>=0`.
If curve is given by parametric equations `x=f(t)` and `y=g(t)` then using substitution rule with `x=f(t)` we have that `dx=f'(t)dt` and since `x` is changing from `a` to `b` then `t` is changing from `alpha=f^(-1)(a)` to `beta=f^(-1)(b)`. It is not always a case, sometimes `t` is changing from `beta` to `alpha`.
To find area in polar coordinates of curve on interval `[a,b]` we use same idea as in calculating area in rectangular coordinates.
So, consider region, that is bounded by `theta=a`, `theta =b` and curve `r=f(theta)`.