Solve the initial value problem

Question

Find the antiderivative that satisfies the condition

f(x) = 8x^3 - 2x^-2 ; F(1) = 5

The answer is supposed to be

2x^4 + 2x^-1 + 1

I am not sure what steps I need to take to get there.

Solve the initial value problem

y'(t) = 3/t +6 ; y(1) = 8

Answer is supposed to be

y(t)=3 ln |t| + 6t +2

I'm not sure how to get there.

Answer 1

F(x) = ∫(8x³ - 2x^-2)dx = 2x⁴ + 2/x + C; F(1) = 2 + 2 + C = 5; C = 1; F(x) = 2x⁴ + 2/x + 1;

y(t) = ∫(3/t + 6)dt = lnItI + 6t + C; y(1) = 3ln1 + 6 + C = = 8; C = 2; y(t) = 3lnItI + 6t + 2

Answer 2

For part a:

f'(x)=8x³ - 2x^-2

To find the antiderivative, think "raise the exponent by 1, multiply by the reciprocal of the new exponent (or divide by the new exponent)".

f(x) = 8(1/4)x⁴ - 2 (-1) x^-1 + C (where C is the constant of integration)

f(x) = 2x⁴ + 2x^-1 + C

To find C use the fact that when x = 1, f(x) = 5

5 = 2(1)⁴ + 2 (1)^-1 + C

5 = 2 + 2 + C

C = 1

For part b, since the antiderivative of 1/t is ln (|t|)

the antiderivative for y'(t) is

y(t) = 3ln(|t|) + 6t + C

When t = 1, y = 8.

8 = 3ln(1) + 6(1) + C

Since ln(1) = 0:

8 = 6 + C

C = 2