An integral is another function in which the original function is a derivative. This function is known as the antiderivative. There is a difference between the definite and indefinite integral which will be explored later.
These are the integral formulas you will need to know for your time in Calculus AB/BC
`color(black)(intkf(x)dx=kintf(x)dx)`
`color(black)(int(f(x)+g(x))dx=intf(x)dx+intg(x)dx)`
`color(black)(intu^ndu=1/(n+1)u^(n+1)+c)` this is the power rule
`color(black)(int(du)/u=lnabs(u)+c)`
`color(black)(inta^udu=a^u/(ln(a))+c)`
`color(black)(inte^udu=e^u+c)`
`color(black)(intcos(u)du=sin(u)+c)`
`color(black)(intsin(u)du=-cos(u)+c)`
`color(black)(intsec^2(u)=tan(u)+c)`
`color(black)(int tan(u)du=lnabs(sec(u))+c)`
`color(black)(intsec(u)tan(u)du=sec(u)+c)`
`color(black)(intsec(u)du=lnabs(sec(u)+tan(u))+c)`
`color(black)(intcsc(u)cot(u)du=-csc(u)+c)`
`color(black)(intcsc(u)du=lnabs(csc(u)-cot(u))+c)`
`color(black)(intcsc^2(u)=-cot(u)+c)`
`color(black)(int(du)/(sqrt(1-u^2))=arcsin(u)+c`
`color(black)(int(du)/(1+u^2)=arctan(u)+c)`
`color(black)(int(du)/(xsqrt(u^2-1))=arcsec(abs(u))+c)`
You can find the antiderivatives of functions using the formulas listed above, but for some problems, you will need some extra tools in order to solve the problems you will encounter on the AP test.
u-Substitution is like trying to reverse the chain rule. If f(g(x)) is the original function, then u is like g(x). For example, you cannot do `color(black)(int((3x^2+4x)(x^3+2x^2)^6)dx)` very easily using any of the rules listed above. In order to get the integrand to match any of the known integral formulas, we have to use u-substitution. We would set `color(black)(u=x^3+2x^2)` so then `color(black)((du)/(dx)=3x^2+4xto du=(3x^2+4x)dx)`. We can now sub everything in for u and we will get `color(black)(int((u)^6)du)`, which looks like the power rule integral formula. After this, you just take the integral to get `color(black)(1/7u^7+c)` which is equivalent to `color(black)(1/7(x^3+2x^2)^7+c).` If you are ever unsure about your answer, if you take the derivative of your answer and you should get the original integrand once again.
Now that was a nice and clean example, and a lot of the time, none of the problems will be as easy as that on the AP test, so here is a harder one to work through.
`color(black)(int(cos(2x+3)dx))`. We should set u=2x+3, and `color(black)((du)/(dx)=2to du=2dx)`, but there is no 2 in the original integrand, so what do you do? You would multiply by 1. You multiply by 1 by multiply the integrand by 2 and the outside of the integral by 1/2, since 2x1/2=1, then you're not really changing anything about the expression. Now you can multiply everything in and you will come out with `color(black)(1/2int(2cos(2x+3)dx))` which is equivalent to `color(black)(1/2intcos(u)du)` which becomes `color(black)(1/2sin(u)+c)`, and when you sub in what u is, you have a final result of `color(black)(1/2sin(2x+3)+c)`
This is essentially u-substitution except that you act that you already have the derivative of one function. The equation is `color(black)(intudv=uv-intvdu)` where u=f(x), du=f'(x)dx, v=g(x), and dv=g'(x)dx. Sometimes, integration by parts may require you to do multiple integration by parts within itself, and in order to make it easier, sometimes you can use the tabular method. You should only really use the tabular method when you have a polynomial term
When you do the tabular method, you take three columns, and you have signs on the first column, u on the second, and dv on the last. On the signs column, you keep alternating signs starting from +, so row 1 is +, row 2 is -, etc. On the u column, you keep taking the derivative until you get to 0, on the dv column, you take the integral of dv, and move all the rows of dv up one. After that, just multiply straight across starting from the second row
Here is an example of integration by parts: `color(black)(lnxdx)` where u=lnx, du=dx/x, dv=1dx, v=x so `color(black)(lnx=xln(x)-intx/xdx)` which is equivalent to `color(black)(xln(x)-x+c)`
Here is an example of the tabular method: `color(black)(intx^5e^(2x)dx)`
| Sign | u | dv |
|---|---|---|
| `color(black)(e^(2x))` | ||
| + | `color(black)(x^5)` | `color(black)(1/2e^(2x))` |
| - | `color(black)(5x^4)` | `color(black)(1/4e^(2x))` |
| + | `color(black)(20x^3)` | `color(black)(1/8e^(2x))` |
| - | `color(black)(60x^2)` | `color(black)(1/16e^(2x))` |
| + | `color(black)(120x)` | `color(black)(1/32e^(2x))` |
| - | `color(black)(120)` | `color(black)(1/64e^(2x))` |
| + | `color(black)(0)` | `color(black)(1/120e^(2x))` |
So after multiply across you should get `color(black)(x^5/2e^(2x)-(5x^4)/4e^(2x)+(20x^3)/8e^(2x)-(60x^2)/16e^(2x)+(120x)/32e^(2x)-120/64e^(2x))`
Sometimes you cannot use u-substitution to complete `color(black)(du/u)` so you have to use partial fractions. In this case, factor the denominator, and separate the factors out into `color(black)(intA/(v) + intB/(w) + ...)`, then A + B + ... should equal what the numerator used. After this, you can use the ln rule to integrate.
Here's an example: `color(black)((x+1)/(x^2+5x+6)dx)`, you cannot use `color(black)((du)/u)` because there is no number that you can multiply by to get the top to be the derivative of the bottom, so instead you factor the bottom into (x+3)(x+2), then separate the integrand into partial fractions of `color(black)(intA/(x+3)+intB/(x+2))`, then you cross multiply so A(x+2) + B(x+3) = x+1. Try to find out what A and B is by making the other one zero, so, for example, set x=-2, and you get A(0) + B(1) = -1, so B = -1, then set x=-3, and you get A(-1) + B(0) = -2, so A=2. Now the integral looks like `color(black)(int2/(x+3)+int-1/(x+2))` which equals `color(black)(2lnabs(x+3)-lnabs(x+2)+c)`
There are two types of improper integrals: The first kind involve one or both limits of integration being infinite, such as `color(black)(int_(-oo)^23x^2dx)`. The other is where the integral has an essential discontinuity in the closed interval [a, b], such as `color(black)(int_(-1)^2 1/(x^3)dx)`
In order to solve the infinite integral, you must add limits to the integral. For example `color(black)(int_0^oo f(x)dx to lim_(ctooo)int_0^c f(x)dx)`. Afterwards, you solve the for the integrals, and the integral converges to a certain finite number when you plug in infinity for c in F(c) - F(a). However, the integral diverges if the integral comes out to be infinity after plugging in infinity for c in F(c) - F(a).
In order to solve for an integral with an essential discontinuity on the closed interval [a, c] with the discontinuity at x=b, then `color(black)(int_a^cf(x)dx)` must be split into `color(black)(int_a^bf(x)dx + int_b^cf(x)dx)` If the integral converges to a finite number, then the integral exists. If it goes to infinity, then the integral diverges.
Here's some practice.
Question 1: What is `color(black)(intx^3+3x^2+8x+4dx)?`
Question 2: What is `color(black)(int(12x)/(x^2)dx)`?
Question 3: What is `color(black)(int9sec(x)tan(x)dx)`?
Question 4: What is `color(black)(int(2cos(x))/(9+sin(x))dx)`?
Click on the up carrot to get the answer.
Answer 1: `color(black)(1/4x^4+x^3+4x^2+4x+c)`
Answer 2: `color(black)(6ln(abs(x^2))+c)`
Answer 3: `color(black)(9tan(x)+c)`
Answer 4: `color(black)(2ln(abs(9+sin(x)))+c)`
Indefinite Integrals are those that do not have limits of integration, this means that you are only finding the antiderivative, which is denotated by a upper case letter for the function. For example, the antiderivative of f(x) is F(x).
Whenever you find an indefinite integral, you always have to have a +c, the reason for this is that the original function, or the derivative of the antiderivative is just the slope of the function, it does not tell you where the function has a point, you can see this in the graph below.
In the graph above, all the functions have the same slope, so they have the same derivative, but when you take the indefinite integral of the derivative, it could be any of the graphs, you do not know unless you plug in a point to find what c in +c will be. By adding +c, we are acknowledging that the antiderivative function can exist at anywhere because the c could be any value until we are given an initial condition.
*an initial condition is something like (1, 2), or some point that is on the antiderivative.
The definition of a definite integral is `color(black)(lim_(ntooo)sum_(k=1)^nf(x_k)(b-a)/n=int_a^bf(x)dx`. This is basically adding up rectangles that extend from the x-axis (lower bound) that have an infintely small width, which essentially returns the area under the curve.
Definite integrals are those that have a lower and upper limit of integration and do not have a +c when integrated, instead it looks like:
`color(black)(int_a^bf(x)=[F(x)]_a^b=F(b)-F(a))`
Definite integrals return the area under the curve between the two limits of integration.
The graph below shows two functions, the red line is the original function and the blue line is the area of the red line from 0 to the x value with respect to the x-axis. You can play around with this graph by clicking on edit this graph and you will be directed to the graph page on Desmos.
Here is another graph that you can play around with. This is a graph of the definite integral where you are able to change the limits of integration. The purple line is the original function and the highlighted part is the area that is being returned by the definite integral.
You can approximate definite integrals through left-hand Riemann sums, right-hand Riemann sums, mid-point Riemann sums, and trapezoidal sums.
Left-Hand Riemann Sums: `color(black)(int_a^bf(x)dx~~f(x_0)xxDeltax + f(x_1)xxDeltax ... f(x_(n-1))xxDeltax)`. This is an overestimate when f(x) is decreasing and an underestimate when f(x) is increasing.
Right-Hand Riemann Sums: `color(black)(int_a^bf(x)dx~~f(x_1)xxDeltax + f(x_2)xxDeltax ... f(x_(n))xxDeltax)`. This is an overestimate when f(x) is increasing and an underestimate when f(x) is decreasing.
Mid-Point Riemann Sums: `color(black)(int_a^bf(x)dx~~f((x_0+x_1)/2)xxDeltax + ... f((x_(n-1)+x_n)/2)xxDeltax)`
Trapezoidal Sums: `color(black)(int_a^bf(x)dx~~(f(x_0)+f(x_1))/2xxDeltax + ... (f(x_(n-1))+f(x_n))/2xxDeltax`
`color(black)(int_a^bf(x)=F(b)-F(a))` where f(x) is the original function and F(x) is the antiderivative. This tells us how to find the area under a curve.
`color(black)(d/(dx)int_a^xf(t)dt=f(x))` when f is continuous on [a, b]. This theorem can be expanded beyond just x in the limits of integration, so in general: `color(black)(d/(dx)int_a^uf(t)dt=f(u)du)`, so for `color(black)(d/(dx)int_a^(x^2)f(t)dt=f(x) xx 2x)`
Here is an example of the second fundamental theorem of calculus. The orange dotted line is thhe original function, the blue line is the antiderivative of the function, and the green line is the derivative of the antiderivative. As you can see, the green line and the orange line are totally equal.
`color(black)(int_a^bkf(x)dx=kint_a^bf(x)dx`
`color(black)(int_a^af(x)dx=0)`
`color(black)(int_a^bf(x)dx=-int_b^af(x)dx`
`color(black)(int_a^cf(x)dx+int_c^bf(x)dx=int_a^bf(x)dx)`
`color(black)(int_a^b(f(x)+g(x))dx=int_a^bf(x)dx+int_a^bg(x)dx)`
`color(black)(int_a^b(f(x)-g(x))dx=int_a^bf(x)dx-int_a^bg(x)dx)`
In order to solve a differential equation there are certain steps that you have to follow.
1. Separate the variables, x's on one side, y's on the other, etc.
2. Integrate and add +c
3. Simplify as much as possible then use the initial condition to find what c is by plugging in the x-values and y-values.
Here's some practice.
Solve the equation `color(black)((dy)/(dx)=8xy+9x^2y)` with the initial solution of (0, 10)?
Click on the up carrot to get the answer.
Answer: `color(black)(y=10e^(4x^2+3x^3))`
In order to find the displacement from time a to b, do `color(black)(int_a^bv(t)dt)`. This is a vector quantity and does include direction.
In order to find the total distance traveled from time a to b, do `color(black)(int_a^babs(v(t))dt)`. This is a scalar quantity and does not include direction.
In order to find the current position of an object at time b, use your initial condition of (a, s(a)) and do `color(black)(s(a)+int_a^bv(t)dt)` where s(t) is position function.
In order to go from velocity function to position function, find the antiderivative of velocity and solve for c using an initial equation. To go from acceleration to velocity, find the antiderivative of acceleration and solve for c using an initial equation.
If the integral `color(black)(int_a^bf(x)dx)` gives the area under f(x) from a to b, and area is just width x height, then if you divide area by width, then you will get the height. If the width is b-a, then the average height or value of a function is `color(black)(1/(b-a)int_a^bf(x)dx)`.
Here's some practice.
Question: What is the average value of `color(black)(3x^2-2)` over the interval [0, 3]?
Click on the up carrot to get the answer.
Answer: 7
In order to find the area under a curve, find out which function is higher than the other and subtract the higher one by the lower one in the integrand.
For example, if f(x) is higher than g(x) then do `color(black)(int_a^b(f(x)-g(x))dx)`
In the graph below, the blue line is y=x and the red line is `color(black)(y=x^3)`
Here's some practice.
Question 1: What is the area under the curve `color(black)(f(x)=sin(x)+3)` over the interval `color(black)([0, 2 pi])`?
Question 2: What is the area enclosed between the curves `color(black)(f(x)=cos^2((pi/2)x)` and `color(black)(g(x)=x^3+3/4)`?
Click on the up carrot to get the answer.
If you wanted the area under the curves from x=-1 to x=1 then you would do `color(black)(int_(-1)^0(x^3-x)dx)+int_0^1(x-x^3)dx)`
`color(black)(Volume = int_a^bA(x)dx)` where A(x) is the area formula and dx is the thickness of the slice. When you do cross-sections like this, you are essentially taking the area of a shape and multiplying it by a thickness to get a volume, except the thickness is extremely infinitesimal small amount and you are taking the area at points like x=2 and x=2.0000000000001, except a smaller difference. By doing this, you make a smooth shape.
If there is only one function then the washer method equation is `color(black)(Volume = piint_a^bR^2dx)` where R is the positive distance between the function and the axis of rotation. If there is two functions then the washer method The washer method equation is: `color(black)(Volume = piint_a^b(R^2-r^2)dx)` where R is the largest positive distance that can be made between the axis of rotation and one of the functions and where r is the smallest positive distance that can be made between the axis of rotation and the other function. This is the best way to get the volume if the function is parallel to the axis of rotation, this means that if you are given y=f(x) and you are rotating around a y=k axis of rotation, you should use washer method. Same thing with x=f(y) and you are rotating around x=k.
Here's some practice.
Question: What is the volume of the lines `color(black)(f(x)=x^2)` and `color(black)(g(x)=x+2)` rotated around y=-2?
Click on the up carrot to get the answer.
Answer: 45.2389 (You can see the work on this graph)
Use this method if the function is perpindicular to the axis of rotation. This means that if you are given y=f(x) and you are rotating around x=k axis of rotation, you should use the shell method, and if you are given x=f(y) and you are rotating around y=k axis of rotation, you should use the shell method.
The equation for shell method is `color(black)(Volume = 2piint_a^br(x)h(x)dx)` where r(x) is the positive distance from the axis of rotation to the representative rectangles and h(x) is the function.
The formula for arc length is `color(black)(int_a^bsqrt(1+((dy)/(dx))^2)dx)`
If you want to know how this equation was derived, Sal Khan made a very good video explaining how arc length is derived, see it here.
Derivatives and integrals are extremely connected throughout calculus. Consider the function f(x), the derivative is f'(x), and second derivative is f''(x), and so on. The antiderivative (with +c being determined) of f''(x) is f'(x), and the antiderivative of f'(x) is f(x), and the antiderivative of f(x) is F(x).
This connection between derivatives and integrals is best seen in the motion problems. Consider the position function s(t), the y-axis is meters, and the x-axis is seconds. When you take the derivaitve, you find the slope, or rise/run, which is meters/seconds, which is what velocity is. And when you take the slope of velocity, which the y-axis is meters/seconds and the x-axis is seconds, you get acceleration which is `color(black)((meters)/((seconds)^2))`.
Then when you take the antiderivative of acceleration, whose y-axis is `color(black)((meters)/((seconds)^2))` and the x-axis is seconds, you find the area by multiplying width and height, where the width is seconds, so you have the result of meters/seconds because `color(black)((meters)/((seconds)^2) xx seconds=(meters)/(seconds))`. Then when you take the antiderivative of velocity, you do meters/seconds multiplied by seconds to get meters, which is position.