Integration

Question

The equation of a curve is y=x⁴+4x+9.

 

(i.)  Find the coordinates of the stationary point on the curve and determine its nature.

(ii.) Find the area of the region enclosed by the curve, the x-axis and the lines x=0 and x=1. 

Answer 1

(i.) A stationary point of differentiable function of one variable is a point of its domain where the derivative is zero.

The given function is differentiable as a polynomial and defined for all reals. Therefore if we can find a point were its derivative vanishes then it will be in the domain. 

Computing derivative: y' = 4x³ + 4. 

Solving: 4x³ + 4 = 0  

Dividing both sides by 4:

         x³ + 1 = 0

 Factoring:

       (x + 1)(x² - x +1) = 0

    x + 1 = 0 or  x² - x +1 = 0, this equation has no real roots

    x = -1

Recall that we need coordinates of stationary point, so we need values of x and y.

For the function y = (-1)⁴ + 4(-1) + 9 = 1 - 4 + 9 = 6.

Ans: (-1, 6) 

 

(ii.)  ∫₀¹ (x⁴ + 4x + 9) dx =(0.2x⁵ + 2x² + 9x)|₀¹ =(0.2(1⁵) + 2(1²) + 9(1)) - (0.2(0⁵) + 2(0²) + 9(0)) = 11.2

Ans: 11.2

 

Answer 2

i.

 

To find the coordinates of the stationary point, find the derivative,

 

y = x⁴+4x+9

y' = 4x³+4

The stationary point is determined when y' = 0

4x³+4 = 0

4(x³+1) = 0

x³ + 1 = 0

factoring this

(x+1)(x²+x+1) = 0

We have only one real 0 here, x = -1.  When x=-1, y=6

 

Pick a point just less than -1 and just above -1.  Let's choose -2 and 0

x    -2     0

y'    -28  4

     dec    inc

So below -1, y is decreasing and above -1, y is increasing.

Therefore (-1,6) is a minimum

This can also be confirmed by the 2nd derivative

y'' = 12x²

evaluate this at x = -1 and y'' = 12 which is positive, confirming our minimum.

 

^ii.

∫₀¹ x⁴+4x+9 dx = x⁵/5 + 4x²/2 + 9x |¹_{0  = 1/5 + 4/2 + 9 -(0+0+0) = 1/5 + 2 + 9 = 11 1/5}