Differentiation

Edexcel IAL WMA12/01 /P2/June 2019/Q10 (Increasing Function Area under Curves)

Figure 1 shows a sketch of part of the curve C with equation y = f(x) where​​ 

fx=36x2 + 2x – 13                x> 0 

Using calculus,​​ 

• ​​ find the range of values of​​ x​​ for which​​ f(x)​​ is increasing,​​ 

(4)

• show that​​ ∫2936x2+2x-13 dx=0

(4)

The point P(2, 0) and the point Q(6, 0) lie on C.​​ 

Given​​ ∫2636x2+2x-13 dx=-8

• ​​ 

• state the value of​​ ∫6936x2+2x-13 dx​​ 

• find the value of the constant k such that​​ ∫2636x2+2x+k dx=0​​ 

(3)

SOLUTION

a-​​ 

To find the range​​ x​​ for which y is increasing, lets find the​​  f'x

fx=36x2 + 2x – 13

f'x= -72 x-3+2

-72x3+2>0

2>72x3

2x3>72

x3>36

x> 363

Hence, fx is increasing when x> 363

b- To​​ show that​​ ∫2936x2+2x-13 dx=0. Lets integrate the equation of​​ f(x).

fx=36x2 + 2x – 13

∫2936x2+2x-13dx=36x-1-1+2x22-13x29

=-36x+x2-13x29 

=-369+81-13 9--362+4-26

=-4+81-117--18+4-26

=-40--40

=-40+40

∫2936x2+2x-13dx=0

Hence, proved.​​ 

c- i- ​​ The value of​​ ∫6936x2+2x-13 dx​​ is the area enclosed in black colour triangle.​​ 

We know that integration gives us the area bounded between the curve and x-axis. Here,​​ ∫2936x2+2x-13 dx=0which we proved in part b, meaning thereby

∫2936x2+2x-13 dx=∫2636x2+2x-13 dx+∫6936x2+2x-13 dx

Where, it is given​​ ∫2636x2+2x-13 dx=-8

0=-8+∫6936x2+2x-13 dx

∫6936x2+2x-13 dx=8

Hence,​​ 

∫6936x2+2x-13 dx=8

c- ii-​​ 

∫2636x2+2x+k dx=0
-36x+x2+kx26=0

Applying limit.

-366+62+6k--362+22+2k=0

-6+26+61+18-4-2k=0

44+4k=0

4k=-44

k=-444

k=-11