This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for \(\displaystyle f'(x) = 45x^{2}+10x+14\), this would look like

\(\displaystyle f(x) = \int (45x^{2}+10x+14)dx\)

which when integrated will equal 

\(\displaystyle 15x^{3}+5x^{2}+14x+C\)

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for \(\displaystyle f'(x) = 32x^{3}+18x^{2}+3x+42\), this would look like

\(\displaystyle f(x) = \int (32x^{3}+18x^{2}+3x+42)dx\)

which when integrated will equal 

\(\displaystyle 8x^{4}+6x^{3}+\frac{3}{2}x^{2}+42x+C\)

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for \(\displaystyle f'(x) = 12x^{3}+12x^{2}+12x+12\), this would look like

\(\displaystyle f(x) = \int (12x^{3}+12x^{2}+12x+12)dx\)

which when integrated will equal 

\(\displaystyle 3x^{4}+4x^{3}+6x^{2}+12x+C\)

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function. 

So, for \(\displaystyle f'(x) = 42x^{6}+\frac{4}{7}x^{3}+76x\), this would look like

\(\displaystyle f(x) = \int (42x^{6}+\frac{4}{7}x^{3}+76x)dx\)

which when integrated will equal 

\(\displaystyle 6x^{7}+\frac{1}{7}x^{4}+38x^{2}+C\)

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

We know that since its a half life question, the exponential used will be:

\(\displaystyle 2^{(-kt)}\) where \(\displaystyle k\) relates to length of the half life. Since the half life is 3 days in this question, \(\displaystyle k=\frac{1}{3}\). 

Since we know how the mass is decaying over time, we simply need to put our \(\displaystyle m\) and \(\displaystyle m_0\) properly such that:

\(\displaystyle m=m_0*2^{-\frac{t}{3}}\).

To check, if you plug \(\displaystyle t=3\), we get that \(\displaystyle m=\frac{1}{2}m_0\), which indicates that we've lost half the mass in 3 days. 

To determine this, we take the derivative of \(\displaystyle y\): 

\(\displaystyle y(t)=e^{-5t}+t\)

\(\displaystyle y'(t)=-5e^{-5t}+1\)

If we set \(\displaystyle y'(t)=y(t)+a(t)\), where \(\displaystyle a(t)\) is a function that sets the two sides equal:

\(\displaystyle -5e^{-5t}+1=e^{-5t}+t+a(t)\)

Notce that if we multiply the right side by \(\displaystyle -5\), and set \(\displaystyle a(t)=-t-\frac{1}{5}\), we get that:\(\displaystyle -5e^{-5t}+1=-5(e^{-5t}+t)-5(-t-\frac{1}{5})=-5e^{-5t}+1\)

Backtracking we get that:

\(\displaystyle y'=-5y+5t+1\)

Although this may seem difficult, since both sides have identical integrals, we can just compare what's underneath the integral. 

\(\displaystyle \int_{a}^{b} \frac{\partial u}{\partial t} dx=-\int_{a}^{b} (\frac{d}{dx} Flux) dx\)

\(\displaystyle \frac{\partial u}{\partial t} =- (\frac{d}{dx} Flux)\)

Moving to the right:

\(\displaystyle \frac{\partial u}{\partial t}+\frac{\mathrm{d} }{\mathrm{d} x}Flux=0\)

We know that:

\(\displaystyle \frac{\mathrm{d}^2 p}{\mathrm{d} t^2}=a(t)\), since acceleration is the 2nd derivative of position. Knowing this we plug this into the formula to get that: 

\(\displaystyle F=m\frac{\mathrm{d}^2 p}{\mathrm{d} t^2}\)

The equation of the line tangent to the curve of f(x) at x=2 will have a slope equal to the instantaneous rate of change of the curve at x=2. In other words, the slope of the tangent line will have a slope equal to the derivative of f(x) defined at x=2, namely f'(2).

To solve this, we have that

\(\displaystyle f(x)=5x^2+3x+8\)  ;  \(\displaystyle f(2)=34\)

\(\displaystyle f'(x)=10x+3\)  ;  \(\displaystyle f'(2)=23\)

To get the equation of the tangent line, we must use the Point-Slope formula utilizing the slope obtained from the derivative and the point of tangency.

The Point-Slope formula is defined as

\(\displaystyle y-y_{1}=m(x-x_{1})\)

in which m is the slope obtained from the derivative, m=23

and (x1,y1) is the point of tangency, (2,34). We then have that

\(\displaystyle y-34=23(x-2)\)

Converting it to slope-intercept form, we have that

\(\displaystyle y=23(x-2)+34=23x-46+34=y=23x-12\)

The equation of the normal line is defined as the line that is perpendicular to the line tangent to the curve at the point of tangency, which occurs at x=2 in this case.

The line perpendicular to the tangent line will have a slope equal to the negative reciprocal of the slope of the tangent line.

If the slope of the tangent line is denoted by the variable n, the normal line will have a slope (m) in which

\(\displaystyle m=-\frac{1}{n}\) 

We begin by defining the point of tangency (x1,y1) and the slope of the tangent line, n. 

\(\displaystyle f(x)=3x^2+x+5\)  ; \(\displaystyle f(2)=20\)

\(\displaystyle f'(x)=6x+1\)  ; \(\displaystyle f'(2)=13=n\)

We then have that the slope of the normal line, m is

\(\displaystyle m=-\frac{1}{13}\)

Utilizing the Point-Slope Formula y-y1=m(x-x1), we have that

\(\displaystyle y-20=-\frac{1}{13}(x-2)\)

Converting to Slope-Intercept form, we get that the slope of the normal line is

\(\displaystyle y=-\frac{1}{13}x+\frac{2}{13}+20=-\frac{1}{13}x+\frac{2}{13}+\frac{260}{13}\)

\(\displaystyle =y=-\frac{1}{13}x+\frac{262}{13}\)