At a relative minimum $\displaystyle (c,f(c))$ of the graph $\displaystyle f(x)$, it will hold that $\displaystyle f'(c) = 0$ and $\displaystyle f''(c) < 0$. 

First, find $\displaystyle f'(x)$. Using the sum rule,

$\displaystyle f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$\displaystyle f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$\displaystyle f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$\displaystyle f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$\displaystyle x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$\displaystyle (x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$\displaystyle x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$\displaystyle (x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$\displaystyle x^{4}+5 = 0$, in which case, $\displaystyle x^{4} = -5$; this equation has no real solutions.

$\displaystyle x^{2}- 6=0$ has two real solutions, $\displaystyle -\sqrt {6}$ and $\displaystyle \sqrt {6}$. 

Now take the second derivative, again using the sum rule:

$\displaystyle f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$\displaystyle f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$\displaystyle f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $\displaystyle \sqrt {6}$ for $\displaystyle x$:

$\displaystyle f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$\displaystyle f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$\displaystyle f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, $\displaystyle f$ has a relative minimum at $\displaystyle x = \sqrt {6}$.

Now. substitute $\displaystyle -\sqrt {6}$ for $\displaystyle x$:

$\displaystyle f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$\displaystyle f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$\displaystyle f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, $\displaystyle f$ has a relative maximum at $\displaystyle x =- \sqrt {6}$.

Because our $\displaystyle \Delta x$ is constant, the left Riemann sum will be

$\displaystyle R_{L}=\Delta x(f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5))$

$\displaystyle R_{L}=0.5\cdot (1+1.75+4+7.75+13+19.75)=23.625$

To use left Riemann sums, we need to use the following formula:

$\displaystyle Area \approx \sum_{i=1}^n f(x_i)\Delta x$.

where $\displaystyle n$ is the number of subintervals, (4 in our problem),

$\displaystyle i$ is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that $\displaystyle i$ will be 1, 2, 3, and then 4)

$\displaystyle f(x_i)$ is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

,$\displaystyle \Delta x$ is the width of each subinterval, which we will determine shortly.

and $\displaystyle \sum_{i=1}^n$ means add all $\displaystyle n$ versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing $\displaystyle f(x_i)\Delta x$, 4 times; 1 time each for $\displaystyle i=1$, $\displaystyle i=2$, $\displaystyle i=3$, and $\displaystyle i=4$. This gives us

$\displaystyle f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

Think of $\displaystyle \Delta x$ as the base of each box, and $\displaystyle f(x_1)$ as the height of the 1st box.

This is basically $\displaystyle Area = (base)(height)$, 4 times, and then added together.

Now we need to determine what $\displaystyle \Delta x, x_1, x_2, x_3,$ and $\displaystyle x_4$ are.

To find $\displaystyle \Delta x$ we find the total length between the beginning and ending x values, which are given in the problem as $\displaystyle x=-2$ and $\displaystyle x=2$. We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, $\displaystyle \Delta x = \frac{b - a}{n}= \frac{(2)- (-2)}{4} = 1$

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from $\displaystyle x=-2$ to $\displaystyle x=2$, we'll just use the smaller one, $\displaystyle -2$, for our first $\displaystyle x_i$.

Now we know that $\displaystyle x_1 = -2$.

To find the next endpoint, $\displaystyle x_2$, just increase the first x by the length of the subinterval, which is $\displaystyle \Delta x = 1$. In other words

$\displaystyle x_2 = x_1 + \Delta x$

$\displaystyle x_2 = -2 + 1$

$\displaystyle x_2 = -1$

Add the $\displaystyle \Delta x$ again to get $\displaystyle x_3$

$\displaystyle x_3 = x_2 + \Delta x$

$\displaystyle x_3 = -1 + 1$

$\displaystyle x_3 = 0$

And repeat to find $\displaystyle x_4$

$\displaystyle x_4 = x_3 + \Delta x$

$\displaystyle x_4 = 0 + 1$

$\displaystyle x_4 = 1$

Now that we have all the pieces, we can plug them in.

$\displaystyle f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

$\displaystyle f(-2)\cdot1 + f(-1) \cdot (1)+ f(0)\cdot (1) + f(1)\cdot (1)$

plug each value into $\displaystyle f(x)=x^4 +1$ and then simplify.

$\displaystyle [(-2)^4+1](1)+[(-1)^4+1](1)+ [(0)^4+1](1)+[(1)^4+1](1)$

$\displaystyle [16 + 1]+ [1 + 1]+ [0+ 1]+ [1 + 1]$

$\displaystyle 17 + 2 + 1 + 2$

$\displaystyle 22$

This is the final answer, which is an approximation of the area under the function.

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f\left ( -2 \right )=5$

$\displaystyle f\left ( 1 \right )=-19$

$\displaystyle f\left ( 4 \right )=5$

$\displaystyle f\left ( 7 \right )=-6$

$\displaystyle f\left ( 10 \right )=-18$

$\displaystyle f\left ( 13 \right )=0$

Although the total interval has a length of $\displaystyle 15$.

Notice the points are equidistant. This distance is our subinterval $\displaystyle 3$ and since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{-2}^{13}f\left ( x \right )\approx 3\left [ (-19)+(5)+(-6)+(-18)+(0)\right ]$

$\displaystyle \int_{-2}^{13}f\left ( x \right )\approx -114$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f(-8)=-13$

$\displaystyle f(-5)=-19$

$\displaystyle f(-2)=6$

$\displaystyle f(1)=-9$

Although the total interval has a length of $\displaystyle 9$ notice the points are equidistant. This distance is our subinterval: $\displaystyle 3$. Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{-8}^{1}f(x)\approx 3\left [ (-19)+(6)+(-9) \right ]$

$\displaystyle \int_{-8}^{1}f(x)\approx -66$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f(-2)=-2$

$\displaystyle f(3)=-15$

$\displaystyle f(8)=14$

$\displaystyle f(13)=15$

$\displaystyle f(18)=-9$

$\displaystyle f(23)=-12$

Although the total interval has a length of $\displaystyle 25$ notice the points are equidistant, with $\displaystyle 5$ intervals between. Each equidistant interval has length: 5

Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{-2}^{23}f(x)\approx 5\left [ (-15)+(14)+(15)+(-9)+(-12) \right ]$

$\displaystyle \int_{-2}^{23}f(x)\approx -35$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f(-7)=18$

$\displaystyle f(-2)=-17$

$\displaystyle f(3)=-16$

$\displaystyle f(8)=-15$

$\displaystyle f(13)=-14$

Although the total interval has a length of $\displaystyle 20$ notice the points are equidistant, with $\displaystyle 4$ intervals between. Each equidistant interval has length: 5

Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{-7}^{13}f(x)\approx 5\left [ (-17)+(-16)+(-15)+(-14) \right ]$

$\displaystyle \int_{-7}^{13}f(x)\approx -310$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f(4)=9$

$\displaystyle f(6)=10$

$\displaystyle f(8)=-18$

$\displaystyle f(10)=15$

$\displaystyle f(12)=18$

Although the total interval has a length of $\displaystyle 8$ notice the points are equidistant, with $\displaystyle 4$ intervals between. Each equidistant interval has length: $\displaystyle 2$. Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{4}^{12}f(x)\approx 2\left [ (10)+(-18)+(15)+(18) \right ]$

$\displaystyle \int_{4}^{12}f(x)\approx 50$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f\left ( -2 \right )=-13$

$\displaystyle f\left ( 6 \right )=-4$

$\displaystyle f\left ( 14 \right )=-15$

$\displaystyle f\left ( 22 \right )=-19$

$\displaystyle f\left ( 30 \right )=18$

$\displaystyle f\left ( 38 \right )=-8$

Although the total interval has a length of $\displaystyle 40$.

Notice the points are equidistant, with $\displaystyle 5$ intervals between them. Each equidistance interval has length: $\displaystyle 8$. Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{-2}^{38}f\left ( x \right )\approx 8\left [ (-4)+(-15)+(-19)+(18)+(-8)\right ]$

$\displaystyle \int_{-2}^{38}f\left ( x \right )\approx -224$

A Riemann sum integral approximation over an interval $\displaystyle \left [ a, b \right ]$ with $\displaystyle n$ subintervals follows the form:

$\displaystyle \int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of $\displaystyle n$ rectangles, each with a base of length: $\displaystyle \frac{b-a}{n}$ and variable heights: $\displaystyle f(x_{i})$, which depend on the function value at $\displaystyle x_{i}$.

In our case, we have function values over an interval:

$\displaystyle f(0)=-19$

$\displaystyle f(5)=14$

$\displaystyle f(10)=2$

$\displaystyle f(15)=15$

Although the total interval has a length of $\displaystyle 15$ notice the points are equidistant, with $\displaystyle 3$ intervals between. Each equidistant interval has length: $\displaystyle 5$. Since we are using the right point of each interval, we disregard the first function value:

$\displaystyle \int_{0}^{15}f(x)\approx 5\left [ (14)+(2)+(15) \right ]$

$\displaystyle \int_{0}^{15}f(x)\approx 155$