Mean Values (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Mean value of a polynomial function

Question: Calculate the mean value of the function $f(x) = 2x^3 + 6x - 1$ in the interval $[2, 5]$.

Solution:

Using the mean value formula with $a = 2$ and $b = 5$:

$\text{Mean value} = \frac{1}{5 - 2} \int_2^5 (2x^3 + 6x - 1) \, dx$

Integrate term by term:

$= \frac{1}{3} \left[ \frac{x^4}{2} + 3x^2 - x \right]_2^5$

Substitute the limits:

$= \frac{1}{3} \left( \frac{625}{2} + 75 - 5 \right) - \frac{1}{3} (8 + 12 - 2)$

$= \frac{1}{3} \left( \frac{625}{2} + 70 \right) - \frac{1}{3}(18)$

$= \frac{243}{2}$

$= 121.5$

Therefore, the mean value is $\frac{243}{2}$ or $121.5$.

Worked Example 2: Mean value with respect to a variable

Question: Calculate the mean value with respect to $t$ of the function $t^2(3t + 1)(t - 2)$ for $1 \leq t \leq 3$.

Solution:

First, expand the brackets:

$t^2(3t + 1)(t - 2) = t^2(3t^2 - 5t - 2)$

$= 3t^4 - 5t^3 - 2t^2$

Apply the mean value formula with $a = 1$ and $b = 3$:

$\text{Mean value} = \frac{1}{3 - 1} \int_1^3 (3t^4 - 5t^3 - 2t^2) \, dt$

$= \frac{1}{2} \left[ \frac{3t^5}{5} - \frac{5t^4}{4} - \frac{2t^3}{3} \right]_1^3$

Substitute the limits:

$= \frac{1}{2} \left( \frac{729}{5} - \frac{405}{4} - 18 \right) - \frac{1}{2} \left( \frac{3}{5} - \frac{5}{4} - \frac{2}{3} \right)$

$= \frac{209}{15}$

$= 13.9 \text{ (to 3 s.f.)}$

Therefore, the mean value is $\frac{209}{15}$ or approximately $13.9$.

Worked Example 3: Mean velocity and mean acceleration

Question: The velocity of a particle after $t$ seconds is given by $v = \frac{1}{8}t\left(9 - \frac{4}{\sqrt{t}}\right)$ m$\cdot$s$^{-1}$.

a) Show that the mean velocity for $1 \leq t \leq 4$ is $2.03$ m$\cdot$s$^{-1}$ to 3 significant figures.

b) Calculate the mean acceleration of the particle over the same time period.

Solution:

Part a:

First, simplify the velocity expression:

$v = \frac{1}{8}t\left(9 - \frac{4}{\sqrt{t}}\right) = \frac{1}{8}\left(9t - 4\sqrt{t}\right)$

$= \frac{9t}{8} - \frac{1}{2}t^{\frac{1}{2}}$

Apply the mean value formula:

$\text{Mean velocity} = \frac{1}{4 - 1} \int_1^4 \left(\frac{9t}{8} - \frac{1}{2}t^{\frac{1}{2}}\right) dt$

$= \frac{1}{3} \left[ \frac{9t^2}{16} - \frac{1}{3}t^{\frac{3}{2}} \right]_1^4$

$= \frac{1}{3}\left(9 - \frac{8}{3}\right) - \frac{1}{3}\left(\frac{9}{16} - \frac{1}{3}\right)$

$= 2.03 \text{ m}\cdot\text{s}^{-1} \text{ to 3 s.f. as required}$

Part b:

Acceleration is the rate of change of velocity: $a = \frac{dv}{dt}$

From $v = \frac{1}{8}t\left(9 - \frac{4}{\sqrt{t}}\right)$, we find:

$\text{Mean acceleration} = \frac{1}{4 - 1} \int_1^4 \frac{dv}{dt} \, dt$

Using the fundamental theorem of calculus:

$= \frac{1}{3} \left[ \frac{1}{8}t\left(9 - \frac{4}{\sqrt{t}}\right) \right]_1^4$

$= \frac{1}{24}(4(9 - 2)) - \frac{1}{24}(1(9 - 4))$

$= 0.958 \text{ m}\cdot\text{s}^{-2}$

Worked Example 4: Finding unknown limits

Question: The mean value of the function $4x + 7$ in the interval $[a, b]$ is $2$ and in the interval $[a, 2b]$ is $3$. Calculate the values of $a$ and $b$.

Solution:

For the first interval $[a, b]$:

$\frac{1}{b - a} \int_a^b (4x + 7) \, dx = 2$

$\frac{1}{b - a} [2x^2 + 7x]_a^b = 2$

$\frac{1}{b - a}(2b^2 + 7b - 2a^2 - 7a) = 2$

$\frac{1}{b - a}(b - a)(2b + 2a + 7) = 2$

$2b + 2a + 7 = 2 \quad \text{...(1)}$

For the second interval $[a, 2b]$:

$\frac{1}{2b - a} \int_a^{2b} (4x + 7) \, dx = 3$

$\frac{1}{2b - a} [2x^2 + 7x]_a^{2b} = 3$

$\frac{1}{2b - a}(8b^2 + 14b - 2a^2 - 7a) = 3$

$\frac{1}{2b - a}(2b - a)(4b + 2a + 7) = 3$

$4b + 2a + 7 = 3 \quad \text{...(2)}$

From equation (1): $2b + 2a + 7 = 2$

From equation (2): $4b + 2a + 7 = 3$

Subtracting equation (1) from equation (2):

$2b = 1$

$b = \frac{1}{2}$

Substituting back into equation (1):

$2\left(\frac{1}{2}\right) + 2a + 7 = 2$

$1 + 2a + 7 = 2$

$2a = -6$

$a = -3$

Therefore, $a = -3$ and $b = \frac{1}{2}$.

Worked Example 5: Mean value with surds

Question: Given that $f(x) = \frac{2x + 1}{\sqrt{x}}$, show that the mean value of $f(x)$ for $x$ in the interval $2 \leq x \leq 8$ is $\frac{31}{9}\sqrt{2}$.

Solution:

First, rewrite the function using index notation:

$f(x) = \frac{2x + 1}{\sqrt{x}} = \frac{2x + 1}{x^{\frac{1}{2}}} = 2x^{\frac{1}{2}} + x^{-\frac{1}{2}}$

Apply the mean value formula:

$\text{Mean value} = \frac{1}{8 - 2} \int_2^8 \left(2x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right) dx$

$= \frac{1}{6} \left[ \frac{4}{3}x^{\frac{3}{2}} + 2x^{\frac{1}{2}} \right]_2^8$

Substitute the limits:

$= \frac{1}{6}\left(\frac{64}{3}\sqrt{2} + 4\sqrt{2}\right) - \frac{1}{6}\left(\frac{8}{3}\sqrt{2} + 2\sqrt{2}\right)$

$= \frac{31}{9}\sqrt{2} \text{ as required}$