Complete the following steps for the given function, interval, and value of \( n \). a. Sketch the graph of the function on the given interval. b. Calculate \( \Delta x \) and the grid points \( x_{0}, x_{1}, \ldots, x_{n} \). c. Illustrate the left and right Riemann sums, and determine which Riemann sum underestimates and which sum overestimates the area under the curve, d. Calculate the left and right Riemann sums. \( f(x)=2 x^{2}+5 \) on \( [1,6] ; n=5 \)

**a. Sketch the graph on the interval** We are given the function \[ f(x)=2x^2+5, \] which is a parabola opening upward with vertex at \((0,5)\). On the interval \([1,6]\), the graph is increasing because the derivative \[ f'(x)=4x \] is positive for \( x>0 \). A sketch would show a curved line starting at \( x=1 \) with \( f(1)=2(1)^2+5=7 \) and rising to \( x=6 \) with \( f(6)=2(6)^2+5=2(36)+5=77 \). **b. Calculate \(\Delta x\) and the grid points** The formula for \(\Delta x\) is \[ \Delta x = \frac{b-a}{n}. \] Here, \( a=1 \), \( b=6 \) and \( n=5 \). Thus, \[ \Delta x = \frac{6-1}{5}=\frac{5}{5}=1. \] The grid points are calculated as follows: \[ x_i = a + i\Delta x \quad \text{for} \quad i=0,1,\ldots,5. \] Thus, - \( x_0 = 1 \) - \( x_1 = 1+1=2 \) - \( x_2 = 1+2=3 \) - \( x_3 = 1+3=4 \) - \( x_4 = 1+4=5 \) - \( x_5 = 1+5=6 \) **c. Illustrate the left and right Riemann sums and determine the underestimator/overestimator** For the left Riemann sum, the height of each rectangle is given by the function value at the left endpoint of each subinterval. In contrast, the right Riemann sum uses the function value at the right endpoint of each subinterval. Since \[ f(x)=2x^2+5 \] is increasing on \([1,6]\), the left Riemann sum will underestimate the area under the curve, and the right Riemann sum will overestimate the area. An illustration: - **Left Riemann Sum:** Rectangles with heights \( f(x_0), f(x_1), f(x_2), f(x_3), f(x_4) \) and width \(\Delta x=1\). - **Right Riemann Sum:** Rectangles with heights \( f(x_1), f(x_2), f(x_3), f(x_4), f(x_5) \) and width \(\Delta x=1\). **d. Calculate the left and right Riemann sums** 1. **Left Riemann Sum** We calculate \( f(x) \) at the points \( x_0, x_1, x_2, x_3, x_4 \): - \( f(1)=2(1)^2+5=2+5=7 \) - \( f(2)=2(2)^2+5=2(4)+5=8+5=13 \) - \( f(3)=2(3)^2+5=2(9)+5=18+5=23 \) - \( f(4)=2(4)^2+5=2(16)+5=32+5=37 \) - \( f(5)=2(5)^2+5=2(25)+5=50+5=55 \) Then, \[ S_{\text{left}}=\Delta x\,[f(1)+f(2)+f(3)+f(4)+f(5)]=1\,(7+13+23+37+55). \] Summing inside: \[ 7+13=20,\quad 20+23=43,\quad 43+37=80,\quad 80+55=135. \] Thus, \[ S_{\text{left}}=135. \] 2. **Right Riemann Sum** Now, calculate \( f(x) \) at the points \( x_1, x_2, x_3, x_4, x_5 \): - \( f(2)=13 \) (as computed above) - \( f(3)=23 \) - \( f(4)=37 \) - \( f(5)=55 \) - \( f(6)=2(6)^2+5=2(36)+5=72+5=77 \) Then, \[ S_{\text{right}}=\Delta x\,[f(2)+f(3)+f(4)+f(5)+f(6)]=1\,(13+23+37+55+77). \] Summing inside: \[ 13+23=36,\quad 36+37=73,\quad 73+55=128,\quad 128+77=205. \] Thus, \[ S_{\text{right}}=205. \] **Summary** - \(\Delta x=1\) with grid points: \( x_0=1,\, x_1=2,\, x_2=3,\, x_3=4,\, x_4=5,\, x_5=6 \). - Left Riemann Sum \( S_{\text{left}}=135 \) (underestimate). - Right Riemann Sum \( S_{\text{right}}=205 \) (overestimate).