If `g(1)=g(2),` then `int_(1)^(2)[f{g(x)}]^(-1)f'{g(x)}g'(x)dx` is equal to

Options

A. 1
B. 2
C. 0
D. 3

Answer

To solve the integral

\[ \int_{1}^{2} f(g(x))^{-1} f'(g(x)) g'(x) \, dx, \]

given that \( g(1) = g(2) \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int_{1}^{2} f(g(x))^{-1} f'(g(x)) g'(x) \, dx. \]
We can recognize that \( f(g(x))^{-1} f'(g(x)) g'(x) \) can be manipulated using substitution.

### Step 2: Use Substitution
Let \( t = f(g(x)) \). Then, we need to find \( dt \):
\[ dt = f'(g(x)) g'(x) \, dx. \]
This implies:
\[ dx = \frac{dt}{f'(g(x)) g'(x)}. \]

### Step 3: Change the Limits
When \( x = 1 \), \( t = f(g(1)) \) and when \( x = 2 \), \( t = f(g(2)) \). Since \( g(1) = g(2) \), we have:
\[ t = f(g(1)) = f(g(2)). \]
Thus, the limits of integration change from \( f(g(1)) \) to \( f(g(2)) \), which are equal.

### Step 4: Substitute into the Integral
Now, substituting into the integral, we get:
\[ \int_{f(g(1))}^{f(g(2))} \frac{1}{t} \, dt. \]
Since the limits of integration are the same, we have:
\[ \int_{f(g(1))}^{f(g(1))} \frac{1}{t} \, dt = 0. \]

### Step 5: Conclusion
Thus, the value of the integral is:
\[ 0. \]

### Final Answer
The answer is \( 0 \).

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