# Questions with Answers on the Second Fundamental Theorem of Calculus

Questions with detailed solutions and explanations on the second theorem of calculus are presented.

## Theorem

The second fundamental theorem of calculus states that if f is a continuous function on an interval I containing a and
F(x) = ax f(t) dt
then F '(x) = f(x) for each value of x in the interval I.

### Question 1

Approximate F'(π/2) to 3 decimal places if
F(x) = 3x 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 '(π/2) = sin( (π/2) 2 ) = 0.624 (3 decimal places)

### Question 2

Let
F(x) = 0x 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) = -1x2 1 / (1 + t 2) dt

Find F'(x)

Solution to Question 3:

• Let u = x2. F is now given by
F(u) = -1u 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 + av(x) f(t) dt
= - au(x) f(t) dt + av(x) f(t) dt

• Since F(x) is the sum of two functions F1 = - au(x) f(t) dt and F2 = av(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))

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