Final, Fall 2021

The average value of a function $f(x)$ over an interval $[a, b]$ is given by the integral of the function over that interval, divided by the length of the interval.

The formula for the average value $f_{avg}$ of a function $f$ on the interval $[a, b]$ is:
$f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx$

For our specific function $g(x) = \sin(x)$ and the interval $[0, \pi]$, we have $a=0$ and $b=\pi$.

So, we need to calculate:
$g_{avg} = \frac{1}{\pi - 0} \int_0^{\pi} \sin(x) dx$




First, we find the antiderivative of $\sin(x)$. The integral of $\sin(x)$ is $-\cos(x)$.

Next, we evaluate the definite integral $\int_0^{\pi} \sin(x) dx$ using the Fundamental Theorem of Calculus.

We evaluate the antiderivative $-\cos(x)$ at the upper limit ($x=\pi$) and subtract its value at the lower limit ($x=0$).




Now we substitute the value of the integral back into the average value formula.

Remember that the length of the interval is $b-a = \pi - 0 = \pi$.

Divide the result of the integral (which is 2) by the length of the interval ($\pi$) to get the average value.


$$\begin{array}{rl}{\displaystyle g_{avg}}& {\displaystyle =\frac{1}{\pi -0}{\int }_0^{\pi }\sin (x)dx}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }{\int }_0^{\pi }\sin (x)dx}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }{[-\cos (x)]}_0^{\pi }}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }((-\cos (\pi ))-(-\cos (0)))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }((-(-1))-(-(1)))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(1-(-1))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(1+1)}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(2)}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{2}{\pi }}}}\end{array}$$
\begin{align*}
g_{avg} &= \frac{1}{\pi - 0} \int_0^{\pi} \sin(x) dx \\
&= \frac{1}{\pi} \int_0^{\pi} \sin(x) dx \\
&= \frac{1}{\pi} \left[ -\cos(x) \right]_0^{\pi} \\
&= \frac{1}{\pi} \left( (-\cos(\pi)) - (-\cos(0)) \right) \\
&= \frac{1}{\pi} \left( (-(-1)) - (-(1)) \right) \\
&= \frac{1}{\pi} \left( 1 - (-1) \right) \\
&= \frac{1}{\pi} (1 + 1) \\
&= \frac{1}{\pi} (2) \\
&= \boxed{\frac{2}{\pi}}
\end{align*}

The expression $(r^2 - x^2)$ appears inside the integral. Recall the equation of a circle centered at the origin with radius $r$: $x^2 + y^2 = r^2$. The upper semi-circle is given by $y = \sqrt{r^2 - x^2}$.

The integral involves $\pi (r^2 - x^2)$, which looks related to the disk method for finding volumes of revolution.

The volume of a solid generated by revolving the area under a curve $y = f(x)$ from $x=a$ to $x=b$ around the x-axis is given by the disk method formula: $V = \int_a^b \pi [f(x)]^2 dx$.

Let's consider the function $y = \sqrt{r^2 - x^2}$. If we revolve the area under this curve from $x=0$ to $x=r$ (which is a quarter-circle) around the x-axis, the volume generated is a hemisphere. Using the disk method, its volume would be:
$V_{hemi} = \int_0^r \pi (\sqrt{r^2 - x^2})^2 dx = \int_0^r \pi (r^2 - x^2) dx$




The integral in the question is $\int_0^r 2\pi (r^2 - x^2) dx$.
We can factor out the 2: $2 \int_0^r \pi (r^2 - x^2) dx$.

Notice that $\int_0^r \pi (r^2 - x^2) dx$ is exactly the integral we identified as the volume of a hemisphere of radius $r$.

Therefore, the given integral is $2 \times V_{hemi}$.
Two hemispheres make a full sphere. So, the integral represents the volume of a sphere of radius $r$.




We can directly calculate the integral to verify. The volume of a sphere is known to be $\frac{4}{3}\pi r^3$. Let's see if the integral evaluates to this.

The calculation shows that the integral indeed equals the formula for the volume of a sphere. Therefore, the correct option is (a).



$$\begin{array}{rl}{\displaystyle I}& {\displaystyle ={\int }_0^{r}2\pi (r^2-x^2)dx}\\ {\displaystyle }& {\displaystyle =2\pi {\int }_0^r(r^2-x^2)dx}\\ {\displaystyle }& {\displaystyle =2\pi {[r^{2}x-\frac{x^3}{3}]}_0^r}\\ {\displaystyle }& {\displaystyle =2\pi ((r^2(r)-\frac{r^3}{3})-(r^2(0)-\frac{0^3}{3}))}\\ {\displaystyle }& {\displaystyle =2\pi (r^3-\frac{r^3}{3}-0)}\\ {\displaystyle }& {\displaystyle =2\pi \left(\frac{3r^3-r^3}{3}\right)}\\ {\displaystyle }& {\displaystyle =2\pi \left(\frac{2r^3}{3}\right)}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{4}{3}\pi r^3}}}\end{array}$$
\begin{align*}
I &= \int_0^r 2\pi (r^2 - x^2) dx \\
&= 2\pi \int_0^r (r^2 - x^2) dx \\
&= 2\pi \left[ r^2x - \frac{x^3}{3} \right]_0^r \\
&= 2\pi \left( (r^2(r) - \frac{r^3}{3}) - (r^2(0) - \frac{0^3}{3}) \right) \\
&= 2\pi \left( r^3 - \frac{r^3}{3} - 0 \right) \\
&= 2\pi \left( \frac{3r^3 - r^3}{3} \right) \\
&= 2\pi \left( \frac{2r^3}{3} \right) \\
&= \boxed{\frac{4}{3}\pi r^3}
\end{align*}

We need the area of the region bounded by the curve $y = \sqrt[3]{x}$, the y-axis (which is the line $x = 0$), and the horizontal line $y = 3$.
The curve $y = \sqrt[3]{x}$ passes through the origin $(0, 0)$.
The region is bounded on the left by $x = 0$, on the right by the curve, below by $y = 0$ (where the curve meets the y-axis), and above by $y = 3$.

Since the region is defined by boundaries involving $y$, it's convenient to integrate with respect to y.




We need to express the right boundary curve in the form $x = f(y)$.

Given $y = \sqrt[3]{x}$, we cube both sides to solve for $x$:
$x = y^3$

The left boundary is the y-axis, which has the equation $x = 0$.




When integrating with respect to y, the area between a right curve $x = f(y)$ and a left curve $x = g(y)$ from $y = c$ to $y = d$ is given by:
$A = \int_c^d [f(y) - g(y)] dy$

In our case:
The right curve is $x_{right} = y^3$.
The left curve is $x_{left} = 0$.
The limits of integration are the y-values defining the region's vertical extent, which are from $y = 0$ to $y = 3$.

So, the integral for the area $A$ is:
$A = \int_0^3 (y^3 - 0) dy$




We evaluate the integral $\int_0^3 y^3 dy$.
The antiderivative of $y^3$ is $\frac{y^4}{4}$.
We evaluate this from $y=0$ to $y=3$.

The result $81/4$ corresponds to option (d).



$$\begin{array}{rl}{\displaystyle A}& {\displaystyle ={\int }_0^3(y^3-0)dy}\\ {\displaystyle }& {\displaystyle ={\int }_0^3{y}^{3}dy}\\ {\displaystyle }& {\displaystyle ={\left[\frac{y^4}{4}\right]}_0^3}\\ {\displaystyle }& {\displaystyle =\left(\frac{3^4}{4}\right)-\left(\frac{0^4}{4}\right)}\\ {\displaystyle }& {\displaystyle =\frac{81}{4}-0}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{81}{4}}}}\end{array}$$
\begin{align*}
A &= \int_0^3 (y^3 - 0) dy \\
&= \int_0^3 y^3 dy \\
&= \left[ \frac{y^4}{4} \right]_0^3 \\
&= \left( \frac{3^4}{4} \right) - \left( \frac{0^4}{4} \right) \\
&= \frac{81}{4} - 0 \\
&= \boxed{\frac{81}{4}}
\end{align*}

Total distance travelled is the integral of the speed, $\int_a^b |v(t)| dt$. This accounts for any changes in direction.
Displacement is the integral of velocity, $\int_a^b v(t) dt$, representing the net change in position.

Since the velocity can be negative, the particle might change direction. We need to integrate the absolute value of velocity to find the total distance.




We need to find where $v(t) = 0$ within the interval $[0, 2]$.
$v(t) = t^2 - 1 = 0$
$(t - 1)(t + 1) = 0$
The solutions are $t = 1$ and $t = -1$.
Only $t = 1$ is within our interval $[0, 2]$. This is where the particle changes direction.

Now we check the sign of $v(t)$ on the subintervals:
On $[0, 1)$: Choose $t = 0.5$. $v(0.5) = (0.5)^2 - 1 = 0.25 - 1 = -0.75$ (Negative)
On $(1, 2]$: Choose $t = 1.5$. $v(1.5) = (1.5)^2 - 1 = 2.25 - 1 = 1.25$ (Positive)




We split the integral at $t = 1$, taking the absolute value on each part.
Distance $D = \int_0^2 |v(t)| dt = \int_0^1 |t^2 - 1| dt + \int_1^2 |t^2 - 1| dt$

Since $v(t)$ is negative on $[0, 1)$, $|t^2 - 1| = -(t^2 - 1) = 1 - t^2$ on this interval.
Since $v(t)$ is positive on $(1, 2]$, $|t^2 - 1| = t^2 - 1$ on this interval.

The integral becomes:
$D = \int_0^1 (1 - t^2) dt + \int_1^2 (t^2 - 1) dt$




We calculate each integral separately.
Integral 1: $\int_0^1 (1 - t^2) dt$
Integral 2: $\int_1^2 (t^2 - 1) dt$

Add the results to find the total distance. The result is $2$, which matches option (c).



$$\begin{array}{rl}{\displaystyle D}& {\displaystyle ={\int }_0^1(1-t^2)dt+{\int }_1^2(t^2-1)dt}\\ {\displaystyle }& {\displaystyle ={[t-\frac{t^3}{3}]}_0^1+{[\frac{t^3}{3}-t]}_1^2}\\ {\displaystyle }& {\displaystyle =((1-\frac{1^3}{3})-(0-\frac{0^3}{3}))+((\frac{2^3}{3}-2)-(\frac{1^3}{3}-1))}\\ {\displaystyle }& {\displaystyle =(1-\frac{1}{3})-0+((\frac{8}{3}-\frac{6}{3})-(\frac{1}{3}-\frac{3}{3}))}\\ {\displaystyle }& {\displaystyle =\left(\frac{2}{3}\right)+((\frac{2}{3})-(-\frac{2}{3}))}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}+(\frac{2}{3}+\frac{2}{3})}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}+\frac{4}{3}}\\ {\displaystyle }& {\displaystyle =\frac{6}{3}}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle 2}}}\end{array}$$
\begin{align*}
D &= \int_0^1 (1 - t^2) dt + \int_1^2 (t^2 - 1) dt \\
&= \left[ t - \frac{t^3}{3} \right]_0^1 + \left[ \frac{t^3}{3} - t \right]_1^2 \\
&= \left( (1 - \frac{1^3}{3}) - (0 - \frac{0^3}{3}) \right) + \left( (\frac{2^3}{3} - 2) - (\frac{1^3}{3} - 1) \right) \\
&= \left( 1 - \frac{1}{3} \right) - 0 + \left( (\frac{8}{3} - \frac{6}{3}) - (\frac{1}{3} - \frac{3}{3}) \right) \\
&= \left( \frac{2}{3} \right) + \left( (\frac{2}{3}) - (-\frac{2}{3}) \right) \\
&= \frac{2}{3} + \left( \frac{2}{3} + \frac{2}{3} \right) \\
&= \frac{2}{3} + \frac{4}{3} \\
&= \frac{6}{3} \\
&= \boxed{2}
\end{align*}

The arc length $L$ of a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the integral:

$L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$

First, we need to find the derivative of $f(x) = y = \frac{2}{3}x^{3/2}$.




Let $f(x) = \frac{2}{3}x^{3/2}$.
Using the power rule, the derivative is:
$f'(x) = \frac{2}{3} \cdot \frac{3}{2} x^{(3/2 - 1)} = 1 \cdot x^{1/2} = \sqrt{x}$

Now, square the derivative:
$[f'(x)]^2 = (\sqrt{x})^2 = x$




Substitute $[f'(x)]^2 = x$ into the arc length formula with the interval $[a, b] = [0, 8]$:
$L = \int_0^8 \sqrt{1 + x} \ dx$




This integral can be solved using a u-substitution.
Let $u = 1 + x$. Then $du = dx$.
We also need to change the limits of integration:
When $x = 0$, $u = 1 + 0 = 1$.
When $x = 8$, $u = 1 + 8 = 9$.

The integral becomes:
$L = \int_1^9 \sqrt{u} \ du = \int_1^9 u^{1/2} \ du$

Now, we find the antiderivative and evaluate. The result $52/3$ corresponds to option (b).



$$\begin{array}{rl}{\displaystyle L}& {\displaystyle ={\int }_1^9{u}^{1\mathrm{/}2}du}\\ {\displaystyle }& {\displaystyle ={\left[\frac{u^{3\mathrm{/}2}}{3\mathrm{/}2}\right]}_1^9}\\ {\displaystyle }& {\displaystyle ={\left[\frac{2}{3}{u}^{3\mathrm{/}2}\right]}_1^9}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(9^{3\mathrm{/}2}-1^{3\mathrm{/}2})}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}((\sqrt{9}{)}^3-1)}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(3^3-1)}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(27-1)}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(26)}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{52}{3}}}}\end{array}$$
\begin{align*}
L &= \int_1^9 u^{1/2} \ du \\
&= \left[ \frac{u^{3/2}}{3/2} \right]_1^9 \\
&= \left[ \frac{2}{3} u^{3/2} \right]_1^9 \\
&= \frac{2}{3} \left( 9^{3/2} - 1^{3/2} \right) \\
&= \frac{2}{3} \left( (\sqrt{9})^3 - 1 \right) \\
&= \frac{2}{3} \left( 3^3 - 1 \right) \\
&= \frac{2}{3} (27 - 1) \\
&= \frac{2}{3} (26) \\
&= \boxed{\frac{52}{3}}
\end{align*}

The interval is $[a, b] = [1, 3]$. We are dividing it into $n$ subintervals.
The width of each subinterval is:
$\Delta x = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n}$




The formula for the right endpoint of the $i$-th subinterval (where $i$ goes from 1 to $n$) is:
$x_i = a + i \Delta x$
Substituting $a = 1$ and $\Delta x = \frac{2}{n}$:
$x_i = 1 + i \left(\frac{2}{n}\right) = 1 + \frac{2i}{n}$




The function is $f(x) = x^2$. We need to evaluate $f(x_i)$:
$f(x_i) = f\left(1 + \frac{2i}{n}\right) = \left(1 + \frac{2i}{n}\right)^2$




The Riemann sum using right endpoints is given by the formula $\sum_{i=1}^n f(x_i) \Delta x$.
Substitute the expressions we found for $f(x_i)$ and $\Delta x$:
$R_n = \sum_{i=1}^n \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n}$

This expression matches option (c).



$$\boxed{{\displaystyle R_n=\sum _{i=1}^n{(1+\frac{2i}{n})}^2\cdot \frac{2}{n}}}$$
\boxed{R_n = \sum_{i=1}^n \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n}}

A function $f$ is odd if $f(-x) = -f(x)$ for all $x$.
A key property for definite integrals of odd functions is:
For any odd continuous function $f$, $\int_{-c}^{c} f(x) dx = 0$ for any $c$ in its domain.
We are asked to evaluate $\int_{-a}^{b} f(x) dx$ where $0 < a < b$.




We can split the interval of integration $[-a, b]$ into two parts in a way that lets us use the property mentioned above. Since $-a < a < b$, we can split the integral at $x = a$:
$\int_{-a}^{b} f(x) dx = \int_{-a}^{a} f(x) dx + \int_{a}^{b} f(x) dx$




The first part of the split integral is $\int_{-a}^{a} f(x) dx$.
Since $f(x)$ is an odd function, we know that:
$\int_{-a}^{a} f(x) dx = 0$




Substitute the result from Step 3 back into the split integral equation from Step 2:
$\int_{-a}^{b} f(x) dx = 0 + \int_{a}^{b} f(x) dx$
$\int_{-a}^{b} f(x) dx = \int_{a}^{b} f(x) dx$

This result matches option (c).



$$\begin{array}{rl}{\displaystyle {\int }_{-a}^{b}f(x)dx}& {\displaystyle ={\int }_{-a}^{a}f(x)dx+{\int }_a^{b}f(x)dx}\\ {\displaystyle }& {\displaystyle =0+{\int }_a^{b}f(x)dx(\text{since }f\text{ is odd})}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle {\int }_a^{b}f(x)dx}}}\end{array}$$
\begin{align*}
\int_{-a}^{b} f(x) dx &= \int_{-a}^{a} f(x) dx + \int_{a}^{b} f(x) dx \\
&= 0 + \int_{a}^{b} f(x) dx \quad (\text{since } f \text{ is odd}) \\
&= \boxed{\int_{a}^{b} f(x) dx}
\end{align*}

Let $h(t) = t^2 e^{-t^6}$. Let's check if it's even or odd.
$h(-t) = (-t)^2 e^{-(-t)^6} = t^2 e^{-t^6} = h(t)$
Since $h(-t) = h(t)$, the integrand $h(t)$ is an even function.




Consider $f(x) = \int_{-x}^{0} h(t) dt$. Let's use the substitution $u = -t$.
This means $t = -u$ and $dt = -du$.
We also change the limits of integration:
When $t = -x$, $u = -(-x) = x$.
When $t = 0$, $u = -0 = 0$.

Substitute into the integral for $f(x)$:
$f(x) = \int_{u=x}^{u=0} h(-u) (-du)$
Since $h(t)$ is even, $h(-u) = h(u)$.
$f(x) = \int_{x}^{0} h(u) (-du)$
Using the property $\int_a^b = -\int_b^a$, we flip the limits and remove the negative sign from $-du$:
$f(x) = \int_{0}^{x} h(u) du$

Now compare this to $g(x) = \int_{0}^{x} h(t) dt$. Changing the variable of integration from $t$ to $u$ doesn't change the value.
Therefore, $f(x) = \int_{0}^{x} h(u) du = g(x)$.

So, $f(x) = g(x)$ for all $x$ in the domain.




Based on the finding that $f(x) = g(x)$:

(a) $f(x) = g(x)$ for all $x \in [0, \infty)$. This is TRUE.

(b) $f'(x) = -g'(x)$ for all $x \in [0, \infty)$. Since $f(x)=g(x)$, their derivatives must be equal: $f'(x)=g'(x)$. Let's find $g'(x)$ using FTC Part 1: $g'(x) = \frac{d}{dx} \int_{0}^{x} t^2 e^{-t^6} dt = x^2 e^{-x^6}$. So $f'(x) = x^2 e^{-x^6}$. The statement $f'(x)=-g'(x)$ would imply $x^2 e^{-x^6} = -x^2 e^{-x^6}$, which is only true if $x=0$. Thus, the statement is FALSE for $x>0$.

(c) $f(x)$ is decreasing and $g(x)$ is increasing on $[0, \infty)$. We found $f'(x) = g'(x) = x^2 e^{-x^6}$. Since $x^2 \ge 0$ and $e^{-x^6} > 0$, both derivatives are non-negative ($\ge 0$). Therefore, both functions are increasing (or non-decreasing). FALSE.

(d) $f(x) = 0$ for all $x \in [0, \infty)$. This implies $g(x)=0$. But $g(x) = \int_{0}^{x} t^2 e^{-t^6} dt$. The integrand $t^2 e^{-t^6}$ is strictly positive for $t \ne 0$. For $x > 0$, $g(x)$ is the integral of a positive function over an interval of positive length, so $g(x) > 0$. FALSE.

(e) $f(x) < 0$ for all $x \in (0, \infty)$. Since $f(x)=g(x)$ and we found $g(x) > 0$ for $x > 0$, this is FALSE.

The relationship between integrals of symmetric functions over symmetric limits is key. For an EVEN function $h(t)$, $\int_{-x}^0 h(t) dt = \int_0^x h(t) dt$. For an ODD function $h(t)$, $\int_{-x}^0 h(t) dt = -\int_0^x h(t) dt$.

The only true statement is (a).

$$\boxed{\text{Statement (a) is true: } f(x) = g(x)}$$

We can split the given series into two separate series using the properties of summation:
$\sum_{n=1}^{\infty} \frac{2^{n+1} + 3^{n-1}}{4^n} = \sum_{n=1}^{\infty} \frac{2^{n+1}}{4^n} + \sum_{n=1}^{\infty} \frac{3^{n-1}}{4^n}$




Let's simplify the first part: $\sum_{n=1}^{\infty} \frac{2^{n+1}}{4^n}$
Rewrite $2^{n+1}$ as $2 \cdot 2^n$.
The series becomes $\sum_{n=1}^{\infty} \frac{2 \cdot 2^n}{4^n} = \sum_{n=1}^{\infty} 2 \left(\frac{2}{4}\right)^n = \sum_{n=1}^{\infty} 2 \left(\frac{1}{2}\right)^n$
This is a geometric series with first term $a$ occurring at $n=1$ and common ratio $r$.
Common ratio $r = \frac{1}{2}$. Since $|r| < 1$, the series converges.
First term (when $n=1$): $a = 2 \left(\frac{1}{2}\right)^1 = 2 \cdot \frac{1}{2} = 1$.

The sum of a convergent geometric series starting from $n=1$ is $\frac{a}{1-r}$, where $a$ is the first term ($n=1$) and $r$ is the common ratio ($|r|<1$).

Sum of the first series = $\frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.




Let's simplify the second part: $\sum_{n=1}^{\infty} \frac{3^{n-1}}{4^n}$
Rewrite $3^{n-1}$ as $3^n \cdot 3^{-1} = \frac{1}{3} \cdot 3^n$.
The series becomes $\sum_{n=1}^{\infty} \frac{\frac{1}{3} \cdot 3^n}{4^n} = \sum_{n=1}^{\infty} \frac{1}{3} \left(\frac{3}{4}\right)^n$
This is a geometric series.
Common ratio $r = \frac{3}{4}$. Since $|r| < 1$, the series converges.
First term (when $n=1$): $a = \frac{1}{3} \left(\frac{3}{4}\right)^1 = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$.

Sum of the second series = $\frac{a}{1-r} = \frac{1/4}{1 - 3/4} = \frac{1/4}{1/4} = 1$.




The sum of the original series is the sum of the two parts:
Total Sum = (Sum of first series) + (Sum of second series) = $2 + 1 = 3$.

This corresponds to option (a).



$$\begin{array}{rl}{\displaystyle S}& {\displaystyle =\sum _{n=1}^{\mathrm{\infty }}\frac{2^{n+1}}{4^n}+\sum _{n=1}^{\mathrm{\infty }}\frac{3^{n-1}}{4^n}}\\ {\displaystyle }& {\displaystyle =\sum _{n=1}^{\mathrm{\infty }}2{\left(\frac{1}{2}\right)}^n+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3}{\left(\frac{3}{4}\right)}^n}\\ {\displaystyle }& {\displaystyle =\left(\frac{{\text{first term}}_1}{1-r_1}\right)+\left(\frac{{\text{first term}}_2}{1-r_2}\right)}\\ {\displaystyle }& {\displaystyle =\left(\frac{2(1\mathrm{/}2{)}^1}{1-1\mathrm{/}2}\right)+\left(\frac{(1\mathrm{/}3)(3\mathrm{/}4{)}^1}{1-3\mathrm{/}4}\right)}\\ {\displaystyle }& {\displaystyle =\left(\frac{1}{1\mathrm{/}2}\right)+\left(\frac{1\mathrm{/}4}{1\mathrm{/}4}\right)}\\ {\displaystyle }& {\displaystyle =2+1}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle 3}}}\end{array}$$
\begin{align*}
S &= \sum_{n=1}^{\infty} \frac{2^{n+1}}{4^n} + \sum_{n=1}^{\infty} \frac{3^{n-1}}{4^n} \\
&= \sum_{n=1}^{\infty} 2 \left(\frac{1}{2}\right)^n + \sum_{n=1}^{\infty} \frac{1}{3} \left(\frac{3}{4}\right)^n \\
&= \left( \frac{\text{first term}_1}{1-r_1} \right) + \left( \frac{\text{first term}_2}{1-r_2} \right) \\
&= \left( \frac{2(1/2)^1}{1 - 1/2} \right) + \left( \frac{(1/3)(3/4)^1}{1 - 3/4} \right) \\
&= \left( \frac{1}{1/2} \right) + \left( \frac{1/4}{1/4} \right) \\
&= 2 + 1 \\
&= \boxed{3}
\end{align*}