Write the Riemann sum to find the area under the graph of the function f (x) = x3 from x = 2 to x = 5.

F (x) = x^3

For calculating Riemann sum to find area under graph from x=2 to x=5,

We can use Riemann's method in 2 parts one is from [0,2] and other [0-5]

So, area = (Riemann Sum of x=0, to x = 5) - (Riemann Sum of x=0, to x = 2)

For Riemann Sum of x=0, to x = 5

The interval [0,5] is divided in subintervals, each of which given a width of 5/n. These are width of Riemann rectangles.

So the Sum = 5/n (5/n) ^3 + ... + 5/n (5*i/n0) ^3 + ... + 5/n (5*n/n) ^3

Taking (5/n) ^4 common

S = (5/n) ^4 * (1 + ... + i^3 + ... + n^3)

S = (5/n) ^4 * [ (n+1) (n+2) / 2]^2

S = (5/n) ^4 * [ (n+1) ^2 * (n+2) ^2]/4

S = 625/n^4 [ (n^2 + 2 + 2n) * (n^2 + 4n + 4) ]/4

S = 625 / (4*n^4) [n^4 + 4n^3 + 4n^2 + 2n^2 + 8n + 8 + 2n^3 + 8n^2 + 8n + ... ]

S = 625/4 [1 + 4/n + 4/n^2 ... ]

if limit is viewed as n->infinity

S = 625/4

Similarly for Riemann Sum of x=0, to x = 2

he interval [0,2] is divided in subintervals, each of which given a width of 2/n. These are width of Riemann rectangles.

So the Sum = 2/n (2/n) ^3 + ... + 2/n (2*i/n0) ^3 + ... + 2/n (2*n/n) ^3

Taking (5/n) ^4 common

S = (2/n) ^4 * (1 + ... + i^3 + ... + n^3)

Solving like above and taking n-> infinity

S = 4

Therefore area under graph f (x) = x^3 from x=2 to x=5 = > 625/4 - 4

=> 152.25