Find f(x) satisfying the given conditions:

Question

1)
f"(x) = 1/x^3
f'(1) = 1/2
f(1) = 0

2)
f"'(x) = sin (x)
f"(0) = 0
f'(0) = 1
f(0) = 0

Answer 1

1 )

f" ( x ) = 1 / x ^ 3 = x ^ ( -3 )

f´ ( x ) = integral of x ^ ( -3 ) dx

f´ ( x ) = x ^ ( - 3 + 1 ) / ( - 3 + 1 ) + C

f´ ( x ) = x ^ ( - 2 ) / ( - 2 ) + C

f´ ( x ) = 1 / x ^ 2 / ( - 2 ) + C

f´ ( x ) = - 1 / ( 2 x ^ 2 ) + C

f´ ( 1 ) = 1 / 2

- 1 / ( 2 * 1 ^ 2 ) + C = 1 / 2

- 1 / ( 2 * 1 ) + C = 1 / 2

- 1 / 2 + C = 1 / 2 Add 1 / 2 to both sides

- 1 / 2 + C + 1 / 2 = 1 / 2 + 1 / 2

C = 1

f´ ( x ) = - 1 / ( 2 x ^ 2 ) + C

f´ ( x ) = - 1 / ( 2 x ^ 2 ) + 1

f ( x ) = integral of f´ ( x ) dx

f ( x ) = integral of [ - 1 / ( 2 x ^ 2 ) ] dx + integral of dx

f ( x ) = - 1 / 2 [ integral of ( 1 / x ^ 2 ) ] dx + integral of dx

f ( x ) = - 1 / 2 [ integral of x ^ ( - 2 ) ] + integral of dx

f ( x ) = - 1 / 2 [ x ^ ( - 2 + 1 ) / ( - 2 + 1 ) ] + integral of dx

f ( x ) = - 1 / 2 [ x ^ ( - 1 ) / ( - 1 ) ] + x + C1

f ( x ) = - 1 / 2 [ - ( 1 / x ) ] + x + C1

f ( x ) = 1 / 2 [ ( 1 / x ) ] + x + C1

f ( x ) = x + 1 / 2 x + C1

f ( 1 ) = 0

1 + 1 / ( 2 * 1 ) + C1 = 0

1 + 1 / 2 + C1 = 0

2 / 2 + 1 / 2 + C1 = 0

3 / 2 + C1 = 0 Subtract 3 / 2 to both sides

3 / 2 + C1 - 3 / 2 = 0 - 3 / 2

C1 = - 3 / 2

So :

f ( x ) = x + 1 / 2 x + C 1

f ( x ) = x + 1 / 2 x - 3 / 2

2 )

f " ( x ) = integral of sin (x) dx

f " ( x ) = - cos ( x ) + C

f " ( 0 ) = 0

- cos ( 0 ) + C = 0

- 1 + C = 0 Add 1 to both sides

- 1 + C + 1 = 0 + 1

C = 1

f " ( x ) = - cos ( x ) + C

f " ( x ) = - cos ( x ) + 1

f´( x ) = integral of f " ( x ) dx

f´( x ) = integral of [ - cos ( x ) + 1 ] dx

f´( x ) = integral of [ - cos ( x ) ] dx + integral of dx

f´( x ) = - integral of [ cos ( x ) ] dx + integral of dx

f´( x ) = - sin ( x ) + x + C1

f´( x ) = x - sin ( x ) + C1

f´( 0 ) = 1

0 - sin ( 0 ) + C1 = 1

0 - 0 + C1 = 1

C1 = 1

f´( x ) = x - sin ( x ) + 1

f ( x ) = integral of f´( x ) dx

f ( x ) = integral of [ x - sin ( x ) + 1 ] dx

f ( x ) = integral of x dx - integral of sin ( x ) + integral of dx

f ( x ) = x ^ 2 / 2 - [ - cos ( x ) ] + x + C2

f ( x ) = x ^ 2 / 2 + cos ( x ) + x + C2

f ( x ) = x ^ 2 / 2 + x + cos ( x ) + C2

f ( 0 ) = 0

0 ^ 2 / 2 + cos ( 0 ) + 0 + C2 = 0

0 + 1 + 0 + C2 = 0

1 + C2 = 0 Subtract 1 to both sides

1 + C2 - 1 = 0 - 1

C2 = - 1

So :

f ( x ) = x ^ 2 / 2 + x + cos ( x ) + C2

f ( x ) = x ^ 2 / 2 + x + cos ( x ) - 1