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π = included in formula booklet β’ π« = not in formula booklet
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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
π = included in formula booklet β’ π« = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
π = included in formula booklet β’ π« = 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
π = included in formula booklet β’ π« = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
π = included in formula booklet β’ π« = not in formula booklet
The area between a curve βf(x)>0β and theΒ βxβ-axis is given by
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbolββ«βΒ and write:
By convention we denote this functionΒ βF:
We can also write
Notice theΒ βdxβΒ under the integral. This tells us which variable we are integrating with respect to - in this case we are reversingΒ βdxdβ.
Since the derivative of a constant is always zero, then if ifΒ βFβ²(x)=f(x),Β thenΒ β(F(x)+C)β²=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 ofΒ βyβΒ orΒ βf(x)βΒ for a givenΒ βx,Β we can determineΒ βCβΒ by plugging inΒ βxβΒ andΒ βy.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
whereΒ βF(x)=β«f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you withΒ ββ«β‘β‘β(β‘)dβ‘.Β 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 anyΒ βa<m<b.
In general, the area enclosed between a curve and theΒ βxβ-axis is given by
since any region below theΒ βxβ-axis hasΒ βf(x)<0,Β but area must always be positive.
This can be done with technology, or by splitting the integral into parts - whereΒ βfβΒ is positive and whereΒ βfβΒ 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 eitherΒ βf(x)>g(x)βΒ orΒ βg(x)>f(x).
The integrals ofΒ βsinβΒ andΒ βcosβΒ 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:
βx1ββΒ is defined forΒ βx<0,Β butΒ βlnxβΒ is not. Specifically:
SoΒ βx1ββΒ is the derivative ofΒ βlnxβΒ and ofΒ βln(βx).
However, we can simplify further. Recall the definition of the absolute value:
Hence, we have
IfΒ βF(x)=β«f(x)dx,Β then
Integrating a composition of functionsΒ βf(g(x))βΒ requires us to divide byΒ βgβ²(x),Β so it is easier to find the anti-derivative of anything of the formΒ βgβ²(x)fβ²(g(x))βΒ by first dividing byΒ βgβ²(x).
In symbols, we use the known fact
and letΒ βu=g(x),Β giving us
an integral we can solve more easily:
Then, we substituteΒ βg(x)βΒ back in to get our desired result ofΒ βkf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are fromΒ βx=aβΒ toΒ βx=b:
We then have two choices:
PlugΒ βx=aβΒ andΒ βx=bβΒ intoΒ βuβΒ to find the bounds in terms ofΒ βu.
PlugΒ βu(x)βΒ back in and use the boundsΒ βaβb.
DisplacementΒ βsβΒ is the change in position between start and end time, whereas distanceΒ βdβΒ 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 betweenΒ βt1β,t2β:
Speed is the magnitude of velocity:
The distance can be found from the velocity using the equation
Since speed is given byΒ ββ£v(t)β£,Β we see that distance is the integral of speed.
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
π = included in formula booklet β’ π« = not in formula booklet
Track your progress:
Don't know
Working on it
Confident
π = included in formula booklet β’ π« = not in formula booklet
The area between a curve βf(x)>0β and theΒ βxβ-axis is given by
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbolββ«βΒ and write:
By convention we denote this functionΒ βF:
We can also write
Notice theΒ βdxβΒ under the integral. This tells us which variable we are integrating with respect to - in this case we are reversingΒ βdxdβ.
Since the derivative of a constant is always zero, then if ifΒ βFβ²(x)=f(x),Β thenΒ β(F(x)+C)β²=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 ofΒ βyβΒ orΒ βf(x)βΒ for a givenΒ βx,Β we can determineΒ βCβΒ by plugging inΒ βxβΒ andΒ βy.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
whereΒ βF(x)=β«f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you withΒ ββ«β‘β‘β(β‘)dβ‘.Β 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 anyΒ βa<m<b.
In general, the area enclosed between a curve and theΒ βxβ-axis is given by
since any region below theΒ βxβ-axis hasΒ βf(x)<0,Β but area must always be positive.
This can be done with technology, or by splitting the integral into parts - whereΒ βfβΒ is positive and whereΒ βfβΒ 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 eitherΒ βf(x)>g(x)βΒ orΒ βg(x)>f(x).
The integrals ofΒ βsinβΒ andΒ βcosβΒ 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:
βx1ββΒ is defined forΒ βx<0,Β butΒ βlnxβΒ is not. Specifically:
SoΒ βx1ββΒ is the derivative ofΒ βlnxβΒ and ofΒ βln(βx).
However, we can simplify further. Recall the definition of the absolute value:
Hence, we have
IfΒ βF(x)=β«f(x)dx,Β then
Integrating a composition of functionsΒ βf(g(x))βΒ requires us to divide byΒ βgβ²(x),Β so it is easier to find the anti-derivative of anything of the formΒ βgβ²(x)fβ²(g(x))βΒ by first dividing byΒ βgβ²(x).
In symbols, we use the known fact
and letΒ βu=g(x),Β giving us
an integral we can solve more easily:
Then, we substituteΒ βg(x)βΒ back in to get our desired result ofΒ βkf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are fromΒ βx=aβΒ toΒ βx=b:
We then have two choices:
PlugΒ βx=aβΒ andΒ βx=bβΒ intoΒ βuβΒ to find the bounds in terms ofΒ βu.
PlugΒ βu(x)βΒ back in and use the boundsΒ βaβb.
DisplacementΒ βsβΒ is the change in position between start and end time, whereas distanceΒ βdβΒ 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 betweenΒ βt1β,t2β:
Speed is the magnitude of velocity:
The distance can be found from the velocity using the equation
Since speed is given byΒ ββ£v(t)β£,Β we see that distance is the integral of speed.