Integrals Problem

Question

∫₄¹⁰ f(x)dx − ∫₄⁹f (x)dx = ∫_a^b f(x)dx

where a= and b=

Answer 1

a= 9, b= 10

integral from 4 to 10 = sum of integrals from 4 to 9 and from 9 to 10

integral of f(x) = x, evaluated from 4 to 10 = x^2/2 100/2 -16/2 = 50-8 = 42

which = sum of integral from 4 to 9 and the integral from 9 to 10

= (81-16)/2= 65/2 plus (100-81)/2 = 19/2. 65/2 + 19/2 = 84/2 = 42