Question

Let f be the function defined by f(x)=3/(2x²−7x+5).

(c) Using the identity that
3/(2x²−7x+5)= [2/(2x-5)] - [1/(x-1)]
evaluate ∫{5,[infinity]} f (x) dx or show that the integral diverges

Answer 1

Step-by-step explanation:

To evaluate the integral ∫[5,∞] f(x) dx, we can use the given identity to rewrite the function f(x) as a difference of fractions:

f(x) = 3/(2x² - 7x + 5) = 2/(2x - 5) - 1/(x - 1).

Now, let's integrate the function f(x) over the given interval [5, ∞]:

∫[5,∞] f(x) dx = ∫[5,∞] (2/(2x - 5) - 1/(x - 1)) dx.

We can integrate each term separately:

∫[5,∞] (2/(2x - 5)) dx = 2∫[5,∞] (1/(2x - 5)) dx.

To evaluate this integral, we can use the u-substitution method:

Let u = 2x - 5, then du/dx = 2, and dx = du/2.

Substituting these values into the integral, we get:

2∫[5,∞] (1/(2x - 5)) dx = 2∫[u(5),∞] (1/u) (du/2)

= ∫[u(5),∞] (1/u) du

= ln|u|∣∣[u(5),∞]

= ln|2x - 5|∣∣[5,∞].

Now let's evaluate the second term of the original integral:

∫[5,∞] (1/(x - 1)) dx.

To evaluate this integral, we can use the u-substitution method again:

Let u = x - 1, then du/dx = 1, and dx = du.

Substituting these values into the integral, we get:

∫[5,∞] (1/(x - 1)) dx = ∫[u(5),∞] (1/u) du

= ln|u|∣∣[u(5),∞]

= ln|x - 1|∣∣[5,∞].

Now, let's evaluate the integral:

∫[5,∞] f(x) dx = ∫[5,∞] (2/(2x - 5) - 1/(x - 1)) dx

= ln|2x - 5|∣∣[5,∞] - ln|x - 1|∣∣[5,∞].

When we evaluate the upper limit of integration (x = ∞), both ln|2x - 5| and ln|x - 1| approach positive infinity. Thus, the integral ∫[5,∞] f(x) dx also approaches positive infinity.

Therefore, we can conclude that the integral ∫[5,∞] f(x) dx diverges.

Answer 2

The integral ∫[5,∞] f(x) dx diverges.

Step-by-step explanation:

To evaluate the given integral ∫{5,[∞]} f(x) dx using the provided identity, we can rewrite the function f(x) as:

f(x) = 3/(2x²−7x+5) = 2/(2x-5) - 1/(x-1)

Now, we can express the integral in terms of this new representation of f(x):

∫{5,[∞]} f(x) dx = ∫{5,[∞]} (2/(2x-5) - 1/(x-1)) dx

We will now evaluate each part separately.

First, let’s consider the integral of 2/(2x-5):

∫{5,[∞]} 2/(2x-5) dx

To solve this, we can use a u-substitution where u = 2x - 5. Then, du = 2dx, and when x = 5, u = 2*5 - 5 = 5. When x approaches infinity, u approaches infinity. The integral becomes:

(1/2) ∫{5,[∞]} du/u

The integral of du/u is ln|u|, so the integral becomes:

(1/2) ln|u| evaluated from 5 to infinity

As u approaches infinity, ln|u| also approaches infinity. Therefore, this part of the integral diverges.

Next, let’s consider the integral of -1/(x-1):

∫{5,[∞]} -1/(x-1) dx

This integral can be evaluated using a simple substitution where v = x - 1. Then, dv = dx, and when x = 5, v = 5 - 1 = 4. When x approaches infinity, v approaches infinity. The integral becomes:

∫{4,[∞]} dv/v

The integral of dv/v is ln|v|, so the integral becomes:

ln|v| evaluated from 4 to ∞

As v approaches infinity, ln|v| also approaches infinity. Therefore, this part of the integral also diverges.

Since both parts of the integral diverge as x approaches infinity, the given integral ∫{5,[∞]} f(x) dx also diverges.