Answer to Question #269755 in Calculus for Thabza

Question

Area Between Curves, Volumes and Average Values of Functions:

1 Calculate the area between the parabola 𝑦=𝑥2 and the line 𝑦=4

2 Derive the formula for the volume of the right-circular cone using a single integral given that 𝑦=−𝑎ℎ(𝑧−ℎ).

3. Find the average value 𝑓(𝑥)̅̅̅̅̅̅ of 𝑓(𝑥)=2𝑥2+3𝑥+3 in the interval [1,4].

Answer 1

1.

$$x^2=4=>x_1=-2, x_2=2$$

$$Area=A=\displaystyle\int_{-2}^{2}(4-x^2)dx=[4x-\dfrac{x^3}{3}]\begin{matrix} 2 \\ -2 \end{matrix}$$

$$=8-\dfrac{8}{3}-(-8+\dfrac{8\dfrac{a}{b}}{3})=\dfrac{32}{3}({units}^2)$$

2.

$$𝑦=−\dfrac{a}{h}(𝑧−ℎ)$$

$$V=\displaystyle\int_{0}^{h}\pi(y(z))^2dz=\displaystyle\int_{0}^{h}\pi(−\dfrac{a}{h}(z-h))^2dz$$

$$=\pi \dfrac{a^2}{h^2}[\dfrac{(z-h)^3}{3}]\begin{matrix} h \\ 0 \end{matrix}=\pi \dfrac{a^2}{3h^2}(0+h^3)=\dfrac{1}{3}\pi a^2 h({units}^3)$$

3.

$$f_{ave}=\dfrac{1}{4-1}=\displaystyle\int_{1}^{4}(2x^2+3x+3)dx$$

$$=\dfrac{1}{3}[\dfrac{2x^3}{3}+\dfrac{3x^2}{2}+3x]\begin{matrix} 4 \\ 1 \end{matrix}$$

$$=\dfrac{1}{3}(\dfrac{128}{3}+24+12-\dfrac{2}{3}-\dfrac{3}{2}-3)$$

$$=\dfrac{49}{2}$$