Integration and Area

Question

Question 1

Question 2

Question 3

Question 4

Answer 1

(4b)

Apply integration by parts  ∫u v dx = u∫v dx −∫u' (∫v dx) dx

Let u = t+1 , v =√t ⇒ u' = 1

Answer 2

(1)

Acceleration of object is a(t) = 3t + 5 m/s²

Acceleration is the rate of change in velocity a(t) = dv/dt

We want to find the velocity v(t)

Given v(0) = 3 m/s

But we have to calculate velocity during the time interval 0 ≤ t ≤ 1

So velocity v(t) = 6.5 m/s

Velocity is the rate of change in distance or displacement v(t) = ds /dt

We have to calculate distance during the time interval 0 ≤ t ≤ 1

So velocity is 6.5 m/s and distance is 6m

Answer 3

(2)

Rate of change of volume is dv/dt  = 100 - ( t - 11)²

Given v = 250 lt when t = 0 means that V(0) = 250

Water in storage tank after 3 days is v(3)

Water in storage tank after 3 days 277 litres

Water in storage tank after 6 days is v(6)

Water in storage tank after 6 days is 448 litres 

Answer 4

(3a)

The area of the region bounded by the the function f(x) = cos x ,the x-axis,

and the vertical lines x = a and x = b :   

The function f(x) = cos x and vertical lines x = π/4 and x = π

      = 0.7071

Area of the region bounded by the the function f(x) = cos x and  x-axis is  0.7071

Answer 5

(4a)

The given integral is 

Answer 6

(4c)

Now spilt the fraction

Answer 7

(4d)

Now spilt the fraction

 

Answer 8

(4e)

Now spilt the terms and perform integration