Question

Use limits to find the area of the region bounded by the graph f(x)=4-2x^3 , the x-axis , and the vertical lines x=0 and x=1

A) 7
B) infinity
C) 7/2
D) 7/4

Answer 1

To find the area of the region bounded by the graph of f(x) = 4-2x^3, the x-axis, and the vertical lines x=0 and x=1, we can use definite integration. The area of the region is 7/4.

Step-by-step explanation:

To find the area of the region bounded by the graph of f(x) = 4-2x^3, the x-axis, and the vertical lines x=0 and x=1, we can use definite integration. First, let's find the x-coordinate of the points where the graph intersects the x-axis by setting f(x) equal to zero:

0 = 4-2x^3

2x^3 = 4

x^3 = 2

x = ∛2

So, the area of the region is given by the integral:

A = ∫0^1 (4-2x^3) dx

After evaluating this integral, we find that the area of the region is 7/4.

Answer 2

C) 7/2

Step-by-step explanation:

The area of the region bounded by the graph of
`f(x)=4-2x^3` , 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.

`\int\limits^1_0 {f(x)} \, dx=\lim_(n\to \infty)\sum_(i=1)^nf(x_i) \triangle x`

Where

`\triangle x=(b-a)/(n)\\ =(1-0)/(n)\\ =(1)/(n)`

Also,

`x_i=a+i\triangle x\\=0+i\cdot (1)/(n) \\=(i)/(n)`

This implies that:

`f(x_i)=4-2((i)/(n))^3=4-(2i^3)/(n^3)`

The formula becomes:

`\int\limits^1_0 {f(x)} \, dx=\lim_(n\to \infty)\sum_(i=1)^n(4-(2i^3)/(n^3) ) (1)/(n)`

`\int\limits^1_0 {f(x)} \, dx=\lim_(n\to \infty)((7)/(2)n-1-(1)/(2n) )/(n)`

`\int\limits^1_0 {4-2x^3} \, dx=\lim_(n\to \infty)(7)/(2)-(1)/(n) -(1)/(2n^2)`

`\int\limits^1_0 {4-2x^3} \, dx=(7)/(2)-(0)-(0)=(7)/(2)`