Pomodoro
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Difficulty
Evaluate the indefinite integral:
Solution
Integrate term by term:
So,
Let
Find $F'(x)$.
Solution
By Part 1 of the Fundamental Theorem of Calculus,
Compute:
Solution
An antiderivative is
Now apply Part 2 of the Fundamental Theorem of Calculus:
Find the average value of
on the interval $[0,3]$.
Solution
Use the average value formula:
Compute the integral:
So the average value is
Evaluate:
Solution
Let
Then the integral becomes
Substitute back:
Evaluate:
Solution
Use integration by parts with
Then
So
Evaluate:
Solution
Write the integrand as partial fractions:
Multiply by $x(x+2)$:
Set $x=0$ to get $1=2A$, so $A=\frac12$. Set $x=-2$ to get $1=-2B$, so $B=-\frac12$.
Thus
Integrate:
Compute:
Solution
Use the identity
Then
Evaluate:
Evaluate the improper integral:
Solution
Write it as a limit:
An antiderivative is
So
The integral converges to $\frac12$.
Use the trapezoidal rule with $n=2$ to approximate
Solution
With $n=2$,
The sample points are $x_0=0$, $x_1=1$, and $x_2=2$.
Evaluate the function:
Apply the trapezoidal rule:
So
Difficulty
Evaluate:
Solution
Let
Change the bounds:
Then
Evaluate:
Solution
Use integration by parts with
Then
So
Now evaluate each part:
and
Therefore
Evaluate:
Solution
Write
Multiply by $x(x+1)^2$:
Expanding gives
Match coefficients:
So
Therefore
Integrate term by term:
Compute:
Solution
Use the substitution
The bounds change as follows:
Also,
So the integral becomes
Find the area between
on the interval $[0,2]$.
Solution
On $[0,2]$, the line $y=2x$ lies above $y=x^2$.
So the area is
Compute:
The region between $y=2$ and $y=x$ for $0 \le x \le 2$ is rotated about the $x$-axis. Find the volume.
Solution
Using washers, the outer radius is $R(x)=2$ and the inner radius is $r(x)=x$.
So
Evaluate:
A thin rod has density
for $0 \le x \le 6$.
Find the mass of the rod.
Solution
Mass is the integral of density:
Compute:
Use Simpson's rule with $n=2$ to approximate
Solution
With $n=2$,
The sample points are $0$, $1$, and $2$.
Evaluate the function:
Apply Simpson's rule:
So
Difficulty
A particle has velocity
for $0 \le t \le 2$.
Find the displacement over that time interval.
Solution
Displacement is the integral of velocity:
An antiderivative is
Evaluate:
So the displacement is $4$.
The region under
from $x=0$ to $x=4$ is rotated about the $y$-axis. Find the volume.
Solution
Using shells, the radius is $x$ and the height is $\sqrt{x}$.
So
Compute:
Therefore
Suppose a random variable has density
for $0 \le x \le 1$.
Find
Solution
Probability is the area under the density:
Compute:
Evaluate:
Solution
This is improper at $x=0$, so write it as a limit:
An antiderivative is
So
Find the geometric area between
and the $x$-axis on $[0,4]$.
Solution
First factor:
So the graph crosses the $x$-axis at $x=1$ and $x=3$.
On $[0,1]$ and $[3,4]$, the function is positive. On $[1,3]$, it is negative.
Thus the geometric area is
An antiderivative is
Evaluate:
So the total area is
Difficulty
Evaluate:
Solution
This is improper at $x=0$, so treat it as a limit:
Use integration by parts with
Then
So
The boundary term at $x=1$ is $0$, and $a^2\ln a \to 0$ as $a\to 0^+$.
Also,
Therefore
Evaluate:
Solution
Write the integral as a limit:
Use the substitution
Then
Compute:
Now let $b\to\infty$:
So the integral converges to $\frac12$.
The region enclosed by
is rotated about the $y$-axis. Find the volume.
Solution
The curves intersect where
so $x=0$ and $x=1$.
Using shells, the radius is $x$ and the height is
Thus
Evaluate:
Compute:
Solution
Use the trig substitution
The bounds become
Also,
So the integral becomes
Use
Then
Evaluate:
Since $\sin(\pi/3)=\frac{\sqrt3}{2}$,
