Questions with detailed solutions on the second theorem of calculus are presented.
The second fundamental theorem of calculus states that if f is a continuous function on an interval I containing a and
F(x) = ∫_{ a}^{x} f(t) dt
then F '(x) = f(x) for each value of x in the interval I.
Question 1:
Approximate F'(Pi/2) to 3 decimal places if
F(x) = ∫_{ 3}^{x} sin(t ^{2}) dt
Solution to Question 1:

Since sin(t ^{2}) is continuous for all real numbers, the second fundamental theorem may be used to calculate F'(x) as follows
F '(x) = sin(x ^{2})

which gives
F '(Pi/2) = sin( (Pi/2) ^{2} ) = 0.624 (3 decimal places)
Question 2:
Let
F(x) = ∫_{ 0}^{x} 5 / (3 + 2 e^{ t}) dt
b) Calculate F'(0)
b) Show that F(1) < F(4)
Solution to Question 2:

a) 5 / (3 + 2 e^{ t}) is continuous, hence the use of the above theorem gives
F'(x) = 5 / (3 + 2 e^{ x})

which gives
F'(0) = 5 / (3 + 2 e^{ 0}) = 1

b) A closer look at F'(x) reveals that F'(x) is positive for all real values of x and therefore function F is an increasing one. Hence since 4 > 1 then
F(4) > F(1) or F(1) < F(4)
Question 3:
Let
F(x) = ∫_{ 1}^{x2} 1 / (1 + t^{ 2}) dt
Find F'(x)
Solution to Question 3:

Let u = x^{2}. F is now given by
F(u) = ∫_{ 1}^{u} 1 / (1 + t^{ 2}) dt

Use the second fundamental theorem to obtain
dF/du = F'(u) = 1 / (1 + u^{ 2})

We now use the chain rule of differentiation to write
F'(x) = dF/dx = dF/du . du/dx = 2x * 1 / (1 + x^{ 4})
Question 4:
Let
F(x) = ∫_{ u(x)}^{v(x)} f(t) dt
where f is continuous everywhere and u and v are continuous functions of x. Express F'(x) in terms of u', v', u, v and f.
Solution to Question 4:

Let a be a real number and write the given integral as the sum (difference) of two integrals as follows
F(x) = ∫_{ u(x)}^{v(x)} f(t) dt
= ∫_{ u(x)}^{a} f(t) dt
+
∫_{ a}^{v(x)} f(t) dt
=  ∫_{ a}^{u(x)} f(t) dt
+
∫_{ a}^{v(x)} f(t) dt

Since F(x) is the sum of two functions F1 =  ò_{ a}^{u(x)} f(t) dt and F2 = ò_{ a}^{v(x)} f(t) dt, then F'(x) is given by
F(x) = F'1(x) + F'2(x)

We now use the second fundamental theorem to write
dF1/du =  f(u)
dF2/dv = f(v)

We now use the chain rule of differentiation to write
F'1(x) = dF1/du * du/dx =  f(u) * du/dx
F'2(x) = dF2/dv * dv/dx = f(v) * dv/dx

Finally F'(x) is given by
F'(x) = F'1(x) + F'2(x)
= v'(x) * f(v(x))  u'(x) * f(u(x))
More on calculus
questions with answers, tutorials and problems .