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Track your progress:
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Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
The area between a curve f(x)>0 and the x-axis is given by
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbol∫ and write:
By convention we denote this function F:
We can also write
Notice the dx under the integral. This tells us which variable we are integrating with respect to - in this case we are reversing dxd.
Since the derivative of a constant is always zero, then if if F′(x)=f(x), then (F(x)+C)′=f(x).
This means that when we integrate, we can add any constant to our result, since differentiating makes this constant irrelevant:
In the same way that constant multiples can pass through the derivative, they can pass through the integral:
And in the same way that the derivative of a sum is the sum of the derivatives:
If we know the value of y or f(x) for a given x, we can determine C by plugging in x and y.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
where F(x)=∫f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you with ∫□□(□)d□. Make sure the variable of your function matches the variable that you take the integral with respect to.
Integrals of the same function with adjacent domains can be merged:
Similarly, the domain of an integral can be split:
for any a<m<b.
In general, the area enclosed between a curve and the x-axis is given by
since any region below the x-axis has f(x)<0, but area must always be positive.
This can be done with technology, or by splitting the integral into parts - where f is positive and where f is negative:
The area enclosed between two curves is given by
This can be done with technology, or by splitting the integral into multiple regions, each having either f(x)>g(x) or g(x)>f(x).
We find the area between a curve and the y-axis by taking an integral with respect to y. Integrals with respect to y have y-value upper and lower bounds and functions in terms of y. Therefore,
where a is the minimum y-value and b is the maximum value, returns the area between the y-axis and a curve f(y):
The integrals of sin and cos are
The anti-derivative of a sum is the sum of the anti-derivatives:
The anti-derivative of a scalar multiple (a constant that does not depend on the integrating variable) can pass through the integral:
x1 is defined for x<0, but lnx is not. Specifically:
So x1 is the derivative of lnx and of ln(−x).
However, we can simplify further. Recall the definition of the absolute value:
Hence, we have
If F(x)=∫f(x)dx, then
Integrating a composition of functions f(g(x)) requires us to divide by g′(x), so it is easier to find the anti-derivative of anything of the form g′(x)f′(g(x)) by first dividing by g′(x).
In symbols, we use the known fact
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=b:
We then have two choices:
Plug x=a and x=b into u to find the bounds in terms of u.
Plug u(x) back in and use the bounds a→b.
Consider the integral
we can break the fraction into two parts:
Then, find A and B using a system of equations. Finally, integrate the two fractions.
To perform integration by parts, we use
Another common way of writing this is
The first step is choosing what to make u and what to make dv. Since we will take the derivative of u, and integrate dv, we generally pick dv that is easy to integrate, and a u that gets simpler when we differentiate it.
We choose u=x and dv=ex, because ex is easy to integrate, and the derivative of x is just 1, which is simpler:
Certain functions, like powers of x, will simplify after repeated differentiation.
Therefore, we may perform repeated integration by parts using such functions as our u knowing that they will eventually yield a simple ∫vdu.
We let u=x2 and v′=e2x because x2 gets simpler when differentiated, and we know how tom integrate e2x:
So
Now we have
So
Cyclical integration by parts are the result of a product of two functions that do not reduce via differentiation. Instead, they "cycle," or eventually return the original integral as the ∫vdu term.
In such cases, integrate by parts until you cycle back to the original integral I. Then, you may solve for I algebraically.
Let u=cos(2x) and v′=ex, then
So
This new integral J can be split by parts, so let u=−2sin2x and v′=ex, then
So
But this last integral is a multiple of the original I, specifically −4I:
From equation (1) we have
Substituting in (2) we find
So
Displacement s is the change in position between start and end time, whereas distance d is the total length of the path taken.
Direction does not matter for distance, which is never negative, but displacement can be negative - usually indicating motion down or to the left.
Acceleration is the rate of change of velocity, which is the rate of change of displacement.
Hence, the integral of acceleration is velocity, and the integral of velocity is displacement.
While we use derivatives to get instantaneous velocity and time, we can also find average velocity and time:
Change in displacement between t1,t2:
Speed is the magnitude of velocity:
The distance can be found from the velocity using the equation
Since speed is given by ∣v(t)∣, we see that distance is the integral of speed.
A curve y=f(x) can be revolved around the x-axis to produce a 3D solid. The following example shows y=2+sinx revolved 2π about the x-axis.
The volume of the resulting solid is given by
The volume of the solid produced by revolving a curve 2π about the y-axis by finding x in terms of y and evaluating
Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
📖 = included in formula booklet • 🚫 = not in formula booklet
The area between a curve f(x)>0 and the x-axis is given by
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbol∫ and write:
By convention we denote this function F:
We can also write
Notice the dx under the integral. This tells us which variable we are integrating with respect to - in this case we are reversing dxd.
Since the derivative of a constant is always zero, then if if F′(x)=f(x), then (F(x)+C)′=f(x).
This means that when we integrate, we can add any constant to our result, since differentiating makes this constant irrelevant:
In the same way that constant multiples can pass through the derivative, they can pass through the integral:
And in the same way that the derivative of a sum is the sum of the derivatives:
If we know the value of y or f(x) for a given x, we can determine C by plugging in x and y.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
where F(x)=∫f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you with ∫□□(□)d□. Make sure the variable of your function matches the variable that you take the integral with respect to.
Integrals of the same function with adjacent domains can be merged:
Similarly, the domain of an integral can be split:
for any a<m<b.
In general, the area enclosed between a curve and the x-axis is given by
since any region below the x-axis has f(x)<0, but area must always be positive.
This can be done with technology, or by splitting the integral into parts - where f is positive and where f is negative:
The area enclosed between two curves is given by
This can be done with technology, or by splitting the integral into multiple regions, each having either f(x)>g(x) or g(x)>f(x).
We find the area between a curve and the y-axis by taking an integral with respect to y. Integrals with respect to y have y-value upper and lower bounds and functions in terms of y. Therefore,
where a is the minimum y-value and b is the maximum value, returns the area between the y-axis and a curve f(y):
The integrals of sin and cos are
The anti-derivative of a sum is the sum of the anti-derivatives:
The anti-derivative of a scalar multiple (a constant that does not depend on the integrating variable) can pass through the integral:
x1 is defined for x<0, but lnx is not. Specifically:
So x1 is the derivative of lnx and of ln(−x).
However, we can simplify further. Recall the definition of the absolute value:
Hence, we have
If F(x)=∫f(x)dx, then
Integrating a composition of functions f(g(x)) requires us to divide by g′(x), so it is easier to find the anti-derivative of anything of the form g′(x)f′(g(x)) by first dividing by g′(x).
In symbols, we use the known fact
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=b:
We then have two choices:
Plug x=a and x=b into u to find the bounds in terms of u.
Plug u(x) back in and use the bounds a→b.
Consider the integral
we can break the fraction into two parts:
Then, find A and B using a system of equations. Finally, integrate the two fractions.
To perform integration by parts, we use
Another common way of writing this is
The first step is choosing what to make u and what to make dv. Since we will take the derivative of u, and integrate dv, we generally pick dv that is easy to integrate, and a u that gets simpler when we differentiate it.
We choose u=x and dv=ex, because ex is easy to integrate, and the derivative of x is just 1, which is simpler:
Certain functions, like powers of x, will simplify after repeated differentiation.
Therefore, we may perform repeated integration by parts using such functions as our u knowing that they will eventually yield a simple ∫vdu.
We let u=x2 and v′=e2x because x2 gets simpler when differentiated, and we know how tom integrate e2x:
So
Now we have
So
Cyclical integration by parts are the result of a product of two functions that do not reduce via differentiation. Instead, they "cycle," or eventually return the original integral as the ∫vdu term.
In such cases, integrate by parts until you cycle back to the original integral I. Then, you may solve for I algebraically.
Let u=cos(2x) and v′=ex, then
So
This new integral J can be split by parts, so let u=−2sin2x and v′=ex, then
So
But this last integral is a multiple of the original I, specifically −4I:
From equation (1) we have
Substituting in (2) we find
So
Displacement s is the change in position between start and end time, whereas distance d is the total length of the path taken.
Direction does not matter for distance, which is never negative, but displacement can be negative - usually indicating motion down or to the left.
Acceleration is the rate of change of velocity, which is the rate of change of displacement.
Hence, the integral of acceleration is velocity, and the integral of velocity is displacement.
While we use derivatives to get instantaneous velocity and time, we can also find average velocity and time:
Change in displacement between t1,t2:
Speed is the magnitude of velocity:
The distance can be found from the velocity using the equation
Since speed is given by ∣v(t)∣, we see that distance is the integral of speed.
A curve y=f(x) can be revolved around the x-axis to produce a 3D solid. The following example shows y=2+sinx revolved 2π about the x-axis.
The volume of the resulting solid is given by
The volume of the solid produced by revolving a curve 2π about the y-axis by finding x in terms of y and evaluating