How would someone solve #16? For **Exercises 15-18**, use the region under the graph of \( f(x) = 2x^3 + 1 \), above the x-axis, and between the vertical lines \( x = 0 \) and \( x = 4 \).15. Use a right sum with \( n = 4 \) to approximate the area of the region.16. Use a left sum with \( n = 4 \) to approximate the area of the region.17. Use a middle sum with \( n = 4 \) to approximate the area of the region.18. Use the limit definition of the area to find the exact value of the area.

Name: How would someone solve #16? For **Exercises 15-18**, use the region u
Uploaded: 2024-03-20T07:06:27.000Z
Channel: Bartleby
Description: How would someone solve #16? For **Exercises 15-18**, use the region under the graph of \( f(x) = 2x^3 + 1 \), above the x-axis, and between the vertical lines \( x = 0 \) and \( x = 4 \).15. Use a right sum with \( n = 4 \) to approximate the area o

Given that, the function is fx=2x3+1 and between the vertical line x=0 and x=4. Therefore, the area of the region is A=∫04fxdx. Use the left sum with n=4 to approximate the area of the region as follows, From the given information, a=0, b=4 and n=4. Obtain ∆x=b-an. ∆x=4-04=1 Divide the interval [0,4] into n=4 subintervals of length ∆x=1 as a=0,11,22,33,4=b. Find the values of fx=2x3+1 for the left end points x=0, 1, 2 and 3 as follows, f0=203+1=0+1=1f1=213+1=2+1=3f2=223+1=16+1=17f3=233+1=54+1=55Substitute the value of f0,f1,f2,f3 and ∆x=1 in ∫abfxdx=∆xfx0+fx1+fx2+⋯+fxn ∫042x3+1dx=∆xfx0+fx1+fx2+fx3=11+3+17+55=176=76 Therefore, the area of the region is 76 units.