Answer:
C) 7/2
Step-by-step explanation:
The area of the region bounded by the graph of [tex]f(x)=4-2x^3[/tex] , the x-axis , and the vertical lines x=0 and x=1 is the integral of f(x) with x=0 and x=1 as limits.
[tex]\int\limits^1_0 {f(x)} \, dx=\lim_{n\to \infty}\sum_{i=1}^nf(x_i) \triangle x[/tex]
Where
[tex]\triangle x=\frac{b-a}{n}\\ =\frac{1-0}{n}\\ =\frac{1}{n}[/tex]
Also,
[tex]x_i=a+i\triangle x\\=0+i\cdot \frac{1}{n} \\=\frac{i}{n}[/tex]
This implies that:
[tex]f(x_i)=4-2(\frac{i}{n})^3=4-\frac{2i^3}{n^3}[/tex]
The formula becomes:
[tex]\int\limits^1_0 {f(x)} \, dx=\lim_{n\to \infty}\sum_{i=1}^n(4-\frac{2i^3}{n^3} ) \frac{1}{n}[/tex]
[tex]\int\limits^1_0 {f(x)} \, dx=\lim_{n\to \infty}\frac{\frac{7}{2}n-1-\frac{1}{2n} }{n}[/tex]
[tex]\int\limits^1_0 {4-2x^3} \, dx=\lim_{n\to \infty}\frac{7}{2}-\frac{1}{n} -\frac{1}{2n^2}[/tex]
[tex]\int\limits^1_0 {4-2x^3} \, dx=\frac{7}{2}-(0)-(0)=\frac{7}{2}[/tex]
