The area A under the graph of fx x3 6 on the interval 2 4

The area A under the graph of f(x) = x^3 - 6 on the interval [2, 4] is approximated using thirty-two subintervals of equal width. If x_k is chosen as the right-hand endpoint of each subinterval, is the resulting approximation greater than, less than, or equal to the real area? The area A under the graph of f(x) = 5x^2 +1 on the interval [-5, -2] is approximated using thirty-two subintervals of equal width. If x_k is chosen as the midpoint of each subinterval, is the resulting approximation

Solution

45) f(x)=x³-6, a=2,b =4,n=32

x=[b-a]/n =(4-2)/32 =1/16=0.0625

area =x*[f(2+0.0625)+f(2+(2*0.0625))+...........+f(4)]

area =0.0625*[(2.0625)³+(2+(2*0.0625))³+(2+(3*0.0625))³+(2+(4*0.0625))³+(2+(5*0.0625))³+(2+(6*0.0625))³+(2+(7*0.0625))³+(2+(8*0.0625))³+(2+(9*0.0625))³+(2+(10*0.0625))³+(2+(11*0.0625))³+(2+(12*0.0625))³+(2+(13*0.0625))³+(2+(14*0.0625))³+(2+(15*0.0625))³+(4)³]

area =0.0625*[306.578125]

area =19.1611328125

real area A=_{[2 to 4]}x³-6 dx

A=_{[2 to 4]}(1/4)x⁴-6x

A=((1/4)4⁴-6*4 )-((1/4)2⁴-6*2 )

A=40-(-8)

A=48

APPROXIMATION LESS TAN ACTUAL AREA