Answer:
[tex]\mathbf{ \int^2_0 \dfrac{x^2}{4}. dx = \dfrac{2}{3}}[/tex]
Step-by-step explanation:
The volume of right pyramid whose base is a square is expressed by the formula:
[tex]V = \int^h_o \dfrac{L^2 *y^2}{h^2}. dy \ or \ V = \int^h_o \dfrac{L^2 *x^2}{h^2}. dx[/tex]
Given that:
height h = 2; &
side (l) = 1
Then;
[tex]V = \int^2_0 \dfrac{(1)^2 *x^2}{(2)^2}. dx[/tex]
[tex]V = \int^2_0 \dfrac{x^2}{4}. dx[/tex]
[tex]V =\dfrac{1}{4} \Big [ \dfrac{x^3}{3} \Big ] ^2_0[/tex]
[tex]V =\dfrac{1}{4} \Big [ \dfrac{(2)^3}{3} -0\Big ][/tex]
[tex]V =\dfrac{1}{4} \Big [ \dfrac{8}{3} -0\Big ][/tex]
[tex]\mathbf{V = \dfrac{2}{3}}[/tex]
