Find \( \int 5 \cos 3x \, dx \) - Leaving Cert Mathematics - Question 5 - 2014

We start with the slope function given by the derivative:

$\frac{dy}{dx} = 2x - 2$

Integrating both sides with respect to ( x ), we get:

$y = x^2 - 2x + c$

To find ( c ), we use the information that the curve cuts the x-axis at ( (-2, 0) ):

At ( x = -2, y = 0 ):

$0 = (-2)^2 - 2(-2) + c \implies 0 = 4 + 4 + c \implies c = -8$

Thus, the equation of the curve is:

$f(x) = x^2 - 2x - 8$

The average value of a function over an interval ([a, b]) is given by:

$\text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx$

For our case, let ( a = 0 ) and ( b = 3 ):

$\text{Average value} = \frac{1}{3 - 0} \int_0^3 (x^2 - 2x - 8) \, dx$

Calculating the integral:

$= \frac{1}{3} \left[ \frac{x^3}{3} - x^2 - 8x \right]_0^3$

Evaluating at the bounds:

$= \frac{1}{3} \left[ \frac{27}{3} - 9 - 24 \right] = \frac{1}{3} \left[ 9 - 9 - 24 \right] = \frac{1}{3}(-24) = -8$

Thus, the average value of ( f ) over the interval is ( -8 ).