integration - calculus

Question

Determine the indefinite integral

2/x + 4/ (x^3)

Then evaluate for 2 to 1

Answer 1

∫(2/x+4/x³)dx = ∫2/x dx + ∫4/x³ dx = 2∫1/x dx + 4∫x^-3 dx = 2ln|x| - 2x^-2 + C = 2ln|x| - 2/x² + C

When you say evaluate from 2 to 1, I'm assuming you mean from 1 to 2. If you really want 2 to 1, then just take the negative of the below:

₁∫²(2/x + 4/x³)dx = (2ln|x} - 2/x²)|₁² = (2ln(2) - 2/2²) - (2ln(1) - 2/1²) = ln(2)² - 1/2 - 0 + 2 = ln(4) + 3/2

Answer 2

Hi Sara!

We are given:

f(x) = 2/x + 4/x³

Which can be written:

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

The problem is asking for the indefinite integral first:

F(x) = ∫ [2/x + 4x^-3]dx

We can pull the constant 2 out in front of the integral:

F(x) = 2∫ [1/x + 2x^-3]dx

The x^-3 term can be integrated easily with reverse power rule, and you may recognize 1/x as the derivative of ln(x), meaning that the antiderivative of 1/x is ln(x).

So, integrating both parts leads to the indefinite integral:

F(x) = 2*[ln(x) - x^-2] + C

Rewriting:

F(x) = 2*[ln(x) - 1/x²] + C

The C is added as an integration constant, as this is an indefinite integral and the starting location of the function is not given.

The next step is to evaluate the indefinite integral from 2 to 1.

In other words, we need to calculate F(1) - F(2), which can be done by subbing in 1 and 2 for the x values:

F(1) = 2*[ln(1) - 1/1²] + C = 2*[0 - 1] + C = C - 2

F(2) = 2*[ln(2) - 1/2²] + C = 2*[ln(2) - 1/4] + C

F(1) - F(2) = (C - 2) - (2*[ln(2) - 1/4] + C) = C - 2 - 2*ln(2) + 1/2 - C = -3/2 - 2*ln(2) ≈ -2.8863

Thus, the exact result is: -3/2 - 2*ln(2)

The approximate result is: -2.8863