Integral Problem

Question

Let

F(x) = integral [From exp x^3 to 1] f(t) dt

Where

f(t) = sqrt(t^2 +5) if t<=2

t + 1 if t>2

a. Evaluate F(ln(cuberoot(3)) - F(ln(cuberoot(2))

b. Is F(x) differentiable? Why?

c. If F(x) is differentiable, find F'(x)

Answer 1

a)

F(x)=∫_exp(x^3)¹f(t) dt

F(ln(3^(1/3)) - F(ln(2^(1/3))=∫₃¹f(t)dt-∫₂¹f(t)dt =∫₃²f(t)dt =

=|f(t)= t+1 for t>2|

=∫₃²(t+1)dt =-∫₂³(t+1)dt = - (t²+t)|₂³ = -12+6=-6

b) F(x) differentiable. The short answer is that it is expressed through and integrable function and is a function of lower limit. Integrate, f(t) is defined for all t (at t=2, f(2)=3). So is F(x). Since f(t) is integrable F(x) has derivative for all x.

c) Derivate of F(x) can be computed according to

∫_g(x)^h(x) f(t) dt = h'(x) · f(h(x)) - g'(x) · f(g(x)),

where h(x)=1 and g(x)=exp(x^3).