Compute the length of the graph of $f(x) = 9 - 4x$ between $x=1$ and $x=5$.
What is our formula for length of curves?
For this problem, we need to find $\displaystyle \int_1^5 \sqrt{1 + f'(x)^2}\ dx$, where $f'(x) = -4$ here.
Thus, we need to find $\displaystyle \int_1^5 \sqrt{5}\ dx$.
You can check the answer here by computing the hypotenuse of a right triangle.
$\displaystyle 4\sqrt{5}$
Compute the length of the graph of $f(x) = \frac{x^4}{16} + \frac{1}{2x^2}$ over the interval $[2,6]$
For our formula, we need $f'(x)$. What is $f'$ here and what does $1 + f'(x)^2$ look like?
$\displaystyle f'(x) = \frac{x^3}{4} - \frac{1}{x^3}$, so $\displaystyle 1 + f'(x)^2 = 1 + \frac{x^6}{16} - \frac{1}{2} + \frac{1}{x^6} = \frac{x^6}{16} + \frac{1}{2} + \frac{1}{x^6}$
In this case $1 + f'(x)^2$ is a perfect square, so $\displaystyle 1 + f'(x)^2 = \left(\frac{x^3}{4} + \frac{1}{x^3}\right)^2$, which makes it easy to work out the arc length integral.
For this answer, we need to compute $\displaystyle \int_2^6 \frac{x^3}{4} + \frac{1}{x^3}\ dx$
$\displaystyle \frac{721}{9} = 80.111 $
Compute the length of the graph of $\displaystyle f(x) = \ln\left(\frac{e^x + 1}{e^x - 1}\right)$ over the interval $[2,4]$.
For this function $\displaystyle f'(x) = \frac{e^x - 1}{e^x + 1} \cdot \frac{(e^x - 1)(e^x) - (e^x + 1)(e^x)}{(e^x - 1)^2} = \frac{-2e^x}{(e^x - 1)(e^x + 1)} = \frac{-2e^x}{e^{2x} - 1}$.
Taking this, we can simplify $1 + f'(x)^2$ to get a perfect square. This simplification gives $\displaystyle 1 + \frac{4e^{2x}}{(e^{2x} - 1)^2} = \frac{e^{4x} - 2e^{2x} + 1 + 4e^{2x}}{(e^{2x} - 1)^2} = \frac{(e^{2x} + 1)^2}{(e^{2x} - 1)^2}$ so that we need to integrate $\displaystyle \int_2^4 \frac{e^{2x} + 1}{e^{2x} - 1}\ dx$.
Using the substitution $u = e^{2x} - 1$ gives $du = 2e^{2x}\ dx = 2(u+1)\ dx$. This gives the integral we need to compute as $\displaystyle \int_2^4 \frac{1}{2}\frac{u+2}{u} \frac{du}{u+1}$, which we can solve by partial fractions.
Since $\frac{u+2}{u(u+1)} = \frac{2}{u} - \frac{1}{u+1}$, we can evaluate this integral from $e^4 - 1$ to $e^8 - 1$.
$\displaystyle \frac{1}{2}\left(2 \ln{e^8 - 1} - 2 \ln{e^4 - 1} - \ln{e^8} + \ln{e^4}\right) = \ln{\left(\frac{e^{8} - 1}{e^4 - 1}\right)} - 2 = 2.0181$
Compute the surface area of the solid of revolution obtained by revolving the graph of $y = \cos{x}$ between $x=0$ and $x = \frac{\pi}{2}$ around the $x$-axis.
What is the formula for computing the surface area of a solid of revolution?
Since the formula is $\displaystyle \int_a^b 2\pi f(x) \sqrt{1 + f'(x)^2}\ dx$, we need to compute $\displaystyle \int_0^{\pi/2} 2\pi \cos(x) \sqrt{1 + \sin^2{x}}\ dx$.
We can evaluate this integral by setting $u = \sin{x}$, so $du = \cos{x}\ dx$. This turns the integral into $\displaystyle 2\pi \int_0^1 \sqrt{1 + u^2}\ du$.
This now requires trig substitution for $u = \tan(\theta)$ and then a reduction formula to get to the final answer here.
$1.1478$
Find the arc length of the curve $y = \sqrt{4 - x^2}$ between $x=-1$ and $x=1$. Compute this using normal integration formulas, and check your answer using geometry.
Find the surface area of the solid of revolution formed by rotating the curve $y = x^2$ between $x=1$ and $x = 5$ around the $x$-axis.