calculus questions?

Question

6. if dy/dx= x^2 + 4, what is the value of y?

a. 2x +c
b. x^3 / 2 + 4x+ c
c. x^3/2 +2x^2 + c
d. x^3/3 + 4x + c


Find the area lying above the x axis and below the parabolic curve y= 4x -x^2

a.8
b. 8 1/3
c. 10 2/3
d. 16

Answer 1

Ok, let's go with those problems :

if dy / dx = x^2 + 4

dy = (x^2 + 4)*dx

integrating, both sides :

y = integrate(x^2 + 4)dx

y = x^3/3 + 4x + C

answer d

Second problem :

Find the area lying above the x axis and below the parabolic curve y= 4x -x^2

That's a parabola opening down, so we must find the intercepts of the parabola and the x axiss :

when y = 0

4x - x^2 = 0 >>>> x(x-4) = 0

x = 0 and x = 4

So we must integrate : 4x - x^2

when x goes from x = 0 to x = 4

integrate(4x - x^2)dx

4*x^2/2 - x^3/3

x from 0 to 4

10 2/3

answer c)

Hope that helps

Answer 2

d)
2) 4x-x^2=0 x= 0 and x=4
A= Integral (0=>4) (4x-x^2)dx =2x^2-1/3 x^3
between 4 and 0
= 32 -64/3 = 32/3= 10 2/3 (c)

Answer 3

#6: Find the integral of x^2+4.

INT[4x-x^2]dx, 0..4