Calculus Integral?

Question

1. y=16-x^2 from 0 to 4 solve the definite integral

2. find y=f(x) given that f"(x) = x^2 f'(0)=7 f(0) = 2

Answer 1

1) Integrating both sides RHS is 16x- (x^3)/3
Put values and we get RHS = 16(4)-4^3 /3 -0-0
= 64-16*4 /3
= 64- 64/3 }
= 64*{1-1/3)
= 64*2/3
= 42.666666666
2) We haven that f''(x) = x^2
Integrating we get f'(x) = x^3 /3 +c
But it is equal to 7 at x = 0
hence c=7
Hence f'(x) +x^3 /3 +7
Integrate again
f(x) = x^4 /12 +7x +c
= 2 at x=0
c=2
f(x) = x^4 /12 + 7x +2

Answer 2

1. y = 16 - x² from 0 to 4 solve the definite integral
∫(16 - x²)dx = 16x - x³/3 | (eval from 0 to 4)
= (16*4 - 64/3) - 0 = 128/3

2. find y=f(x) given that f"(x) = x² f'(0)=7 f(0) = 2
f'(x) = ∫f''(x)dx = ∫x²dx = x³/3 + C = x³/3 + 7
y = f(x) = ∫f'(x)dx = ∫{x³/3 + 7}dx = x^4/12 + 7x + C
y = f(x) = x^4/12 + 7x + 2

Answer 3

Do your own homework.