the integral from 2 to 3 of 1 divided by the quantity 3 times x minus 2, dx is equal to the integral from 4 to 7 of 1 divided by u, du
True?
False?
The answer is False
Well, if we let u = 3x - 2, then du would be equal to 3dx. So, if we take the integral from 2 to 3, the bounds would become 4 to 7. And if we substitute u = 3x - 2, the integrand would become 1/u.
So, the statement is true! They are indeed equal. But don't worry, I won't divide you into pieces to get the answer.
True.
To verify this, let's evaluate both integrals step by step.
1. Let's start with the left-hand side integral: ∫(2 to 3) [1 / (3x - 2)] dx.
2. To evaluate this integral, we can use u-substitution.
Let u = 3x - 2, then du = 3 dx.
Rearranging the equation and solving for dx, we get dx = du / 3.
3. Substituting these into the integral, we have ∫[2 to 3] (1/u)(du/3).
4. Simplifying, we get (1/3) ∫[2 to 3] du/u.
5. Taking the antiderivative of (1/u) with respect to u, we get ln|u|.
6. Evaluating the integral from 2 to 3, we have (1/3)[ln|3x - 2|] evaluated from 2 to 3.
7. Plugging in the upper and lower limits of integration, we get (1/3)[ln|3(3) - 2| - ln|3(2) - 2|].
8. Further simplifying, we have (1/3)[ln|7| - ln|4|].
9. Using logarithmic properties, we can rewrite this as (1/3)[ln(7/4)].
10. Finally, evaluating, we get (1/3)ln(7/4), which is the value of the left-hand side integral.
11. Now, let's consider the right-hand side integral: ∫(4 to 7) (1/u) du.
12. Evaluating this integral directly by taking the antiderivative of (1/u), we get ln|u| evaluated from 4 to 7.
13. Plugging in the upper and lower limits of integration, we have ln|7| - ln|4|.
14. Simplifying, we get ln(7/4), which is the value of the right-hand side integral.
15. Comparing the two integrals, we have (1/3)ln(7/4) = ln(7/4), which is true.
Therefore, the statement is true.
To determine whether the statement is true or false, we need to evaluate both integrals and compare their values.
Let's start by evaluating the integral on the left side:
∫(2 to 3) 1 / (3x - 2) dx
To do this, we can use a method called substitution. Let u = 3x - 2, then du = 3 dx. Rearranging the equation, we can solve for dx: dx = du / 3.
The limits of integration need to be adjusted as well. When x = 2, u = 3(2) - 2 = 4. When x = 3, u = 3(3) - 2 = 7.
Now, we can rewrite the original integral using the new variable u and adjusted limits:
∫(4 to 7) 1/u (du/3)
Simplifying the expression, we have:
(1/3) ∫(4 to 7) 1/u du
Evaluating this integral, we get:
(1/3) ln|u| (evaluated from 4 to 7)
Substituting back u = 3x - 2:
(1/3) ln|3x - 2| (evaluated from 4 to 7)
Next, let's evaluate the integral on the right side:
∫(4 to 7) 1/u du
This integral is straightforward and evaluates to:
ln|u| (evaluated from 4 to 7)
Now, let's compare the two integrals:
(1/3) ln|3x - 2| (evaluated from 4 to 7) vs. ln|u| (evaluated from 4 to 7)
As we can see, the two integrals are not equal. The integral on the left side involves the natural logarithm of a more complex function (3x - 2), while the integral on the right side simply involves the natural logarithm of u. Thus, the statement is false.
I just did the actual question to see if it was true or false.
It was easy to integrate using simple ln
let me know what you found.