Question

1. Find the area between the curve y= x^2-3x from x=0 to x=5.

Answer 1

Area $=\int_0^5 (x^2-3x)dx$

$=\int_0^3(3x-x^2)dx+\int_3^5(x^2-3x)dx$

$=[3x^2/2-x^3/3]_0^3+[x^3/3-3x^2/2]_3^5$

$=(27/2-9)+(5^3/3-3*5^2/2)-(9-27/2)$

$=27-18+(125/3-75/2)$

$=(25/6)+9$

$=79/6$

Answer: $\frac{79}{6}.$