Question: Integrate f(x)=8x+1 between x=12 and x=17.
Upper_bound = 17
Lower_bound - 12
F(x) = INT [f(x)] dx from lower_bound to upper_bound
F(x) = INT [8x + 1] dx
F(x) = INT [8x] dx + INT [1] dx
F(x) = INT [8x] dx + INT [1] dx
Pull out the constant multiple and turn 1 into x^0 = 1
F(x) = 8*INT [x^1] dx + INT [x^0] dx
But INT [x^n] dx = [1/(n+1)]* x^(n+1) GENERAL FORMULA: Power Rule
So F(x) = 8*INT [x^1] dx + INT [x^0] dx
F(x) = 8*(1/2) x^2 + (1/1) * x^1
F(x) = 4x^2 + x GENERAL SOLUTION
The Fundamental Theorem of Calculus states
INT [f(x)] dx = F(upper_bound) - F(lower_bound)
= F(17) - F(12)
= [4x^2 + x] for x = 17 and - [4x^2 + x] for x = 12
= [4[17]^2 + 17] - [4[12]^2 + 12]
= [4*289 + 17] - [4*144 + 12]
= [1156 + 17] - [576 + 12]
= [1173] - [588]
= 585
Therefore, the are under the curve is 585 cubic units
