Application of diff and int

i) dV/dt = t^(3/2)

V = ∫ t^(3/2) dt

= (2/5)t^(5/2) + C

when t = 0 V = 0; ==> C = 0

V = (2/5)t^(5/2) m³

ii) If V = 100

100 = (2/5)t^(5/2)

250 = t^(5/2)

ln(100) = (5/2)ln(t)

t = 6.31 minutes

2. y = -x² -x + 3 ; y = x + 3

x + 3 = -x² -x + 3

x² + x = 0

x(x+1)=0 x=-1 or x=0

when x = 0 sub into y = x + 3 ; y = 3

A is (0,3)

when x = -1 sub into y = x + 3 ; y = 2

B is (-1,2)

The required area is ∫ (-1 to 0) [ C - L ] dx

= ∫ (-1 to 0) [ (-x² -x + 3) - (x + 3) ] dx

= ∫ (-1 to 0) [-x² -2x]dx

= [-(1/3)x³ - x²] 0,-1

= [-(1/3)0³ - 0²] - [-(1/3)(-1)³ - (-1)²]

= 0 - [(1/3) - 1] = -[-(2/3)] = (2/3) square unit