Find f : antiderivative?

Question

1) f''(x)=2+x^3+x^6

2) f'(x)=2x-3/x^4, x>0, f(1)=3

3) f''(x)=4-6x-40x^3, f(0)=3, f'(0)=1


Thanks for the help!! this chapter is too confusing!

Answer 1

1)

f''(x) = 2 + x^3 + x^6

Use the reverse power rule. Remember that the antiderivative of x^n is (1/(n + 1))x^(n + 1)

f'(x) = 2x + (1/4)x^4 + (1/7)x^7 + C

Taking the antiderivative one more time,

f(x) = 2(1/2)x^2 + (1/4)(1/5)x^5 + (1/7)(1/8)x^8 + Cx + D

Simplifying,

f(x) = x^2 + (1/20)x^5 + (1/56)x^8 + Cx + D

2) f'(x) = (2x - 3)/x^4, x > 0, f(1) = 3

First off, Split this into two fractions, both over x^4.

f'(x) = (2x)/x^4 - 3/x^4

Reduce each fraction.

f'(x) = 2/x^3 - 3/x^4

Move each term in the denominator up to the numerator. Doing so would make the powers negative.

f'(x) = 2x^(-3) - 3x^(-4)

Again, we use the reverse power rule. Taking the antiderivative,

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

Simplifying,

f(x) = -x^(-2) + x^(-3) + C

But f(1) = 3, and by definition,
f(1) = -1^(-2) + (1)^(-3) + C
f(1) = -1 + 1 + C
f(1) = C

Equating our given with our evaluation, f(1) = C = 3.

Plugging C = 3 into our general solution,

f(x) = -x^(-2) + x^(-3) + 3

3) f''(x) = 4 - 6x - 40x^3, f(0) = 3, f'(0) = 1

Same rules as above.

f'(x) = 4x - 6(1/2)x^2 - 40(1/4)x^4 + C
f'(x) = 4x - 3x^2 - 10x^4 + C

f(x) = 4(1/2)x^2 - 3(1/3)x^3 - 10(1/5)x^5 + Cx + D
f(x) = 2x^2 - x^3 - 2x^5 + Cx + D

All we have to do is solve for C and D.

f(0) = 2(0)^2 - (0)^3 - 2(0)^5 + C(0) + D = 3
0 - 0 - 0 + 0 + D = 3
D = 3

f'(0) = 4(0) - 3(0)^2 - 10(0)^4 + C = 1
0 - 0 - 0 + C = 1
C = 1

C = 1, D = 3, and our final solution is

f(x) = 2x^2 - x^3 - 2x^5 + x + 3

Answer 2

1) f''(x) = 2 + x^3 + x^6
f'(x) = C1 + 2x + (1/4)x^4 + (1/7)x^7
f(x) = C2 + C1x + x^2 + (1/20)x^5 + (1/56)x^8

2) f'(x) = 2x - 3/x^4, x>0, f(1)=3
f(x) = C + x^2 + 1/x^3
3 = C + 1 + 1
C = 1
f(x) = 1 + x^2 + 1/x^3

3) f''(x) = 4 - 6x - 40x^3, f(0)=3, f'(0)=1
f'(x) = 1 + 4x - 3x^2 - 10x^4
f(x) = 3 + 2x^2 - x^3 - 2x^5