Answer :
Answer:
Average value: 2.30305455955
z ≈ 4.29945519514
Step-by-step explanation:
You want the value of z such that f(z) is the average value of f(x) = √(x+1) on the interval [1, 8].
Average value
The average value is the integral of the function, divided by the width of the interval.
[tex]\displaystyle\overline{y}=\dfrac{1}{x_2-x_1}\int_{x_1}^{x_2}{f(x)}\,dx=\dfrac{1}{7}\int_1^8{\sqrt{x+1}}\,dx\\\\\\\overline{y}=\dfrac{2}{21}\left((8+1)^\frac{3}{2}-(1+1)^\frac{3}{2}\right)=\dfrac{2(27-2\sqrt{2})}{21}\\\\\\\boxed{\overline{y}\approx 2.30305455955}[/tex]
The corresponding value of z so that f(z) = y is ...
[tex]z=\overline{y}^2-1\\\\\boxed{z \approx 4.29945519514}[/tex]
The average value of f(x)= √(x+1) on 1,8 is [tex]\frac{18}{7} - \frac{4\sqrt{2}}{21}[/tex] . For [tex]\[ z = \frac{2948 - 432\sqrt{2}}{441} - 1 \][/tex] on 1,8 does this average value equal f(z).
1. Integral calculation:
- [tex]\[ \int_1^8 \sqrt{x + 1} \, dx \][/tex]
- Using the substitution u = x + 1, du = dx:
[tex]\[ \int_2^9 \sqrt{u} \, du = \int_2^9 u^{1/2} \, du = \left. \frac{2}{3} u^{3/2} \right|_2^9 = \frac{2}{3} \left( 9^{3/2} - 2^{3/2} \right) \][/tex]
- Since [tex]\( 9^{3/2} = 27 \)[/tex] and [tex]\( 2^{3/2} = 2\sqrt{2} \)[/tex]:
[tex]\[ \frac{2}{3} \left( 27 - 2\sqrt{2} \right) = \frac{54 - 4\sqrt{2}}{3} = 18 - \frac{4\sqrt{2}}{3} \][/tex]
2. Average value calculation:
- Average value = [tex]\frac{1}{8 - 1} \int_1^8 \sqrt{x + 1} \, dx = \frac{1}{7} \left( 18 - \frac{4\sqrt{2}}{3} \right) = \frac{18}{7} - \frac{4\sqrt{2}}{21}[/tex].
3. Finding z such that f(z) equals the average value:
[tex]\[ \sqrt{z + 1} = \frac{18}{7} - \frac{4\sqrt{2}}{21} \][/tex]
- Squaring both sides:
[tex]\[ z + 1 = \left( \frac{18}{7} - \frac{4\sqrt{2}}{21} \right)^2 = \left( \frac{54 - 4\sqrt{2}}{21} \right)^2 = \frac{(54 - 4\sqrt{2})^2}{21^2} \][/tex]
- Calculate [tex]\( (54 - 4\sqrt{2})^2 \)[/tex]:
[tex]\[ (54 - 4\sqrt{2})^2 = 54^2 - 2 \cdot 54 \cdot 4\sqrt{2} + (4\sqrt{2})^2 = 2916 - 432\sqrt{2} + 32 = 2948 - 432\sqrt{2} \][/tex]
- So:
[tex]\[ z + 1 = \frac{2948 - 432\sqrt{2}}{441} \][/tex]
- Finally, solving for z:
[tex]\[ z = \frac{2948 - 432\sqrt{2}}{441} - 1 \][/tex].
