If f(x) is continuous function , then

Options

A. ` int_9^16 f(x)dx = 7f(c)` for some 'c' lying between 9 and 16
B. ` int_9^16 f(x)dx` = ` int_tan9^tan16 (f(tan^(-1) t))/(1+ t^(2))dt`
C. ` int_9^16 f(x)dx = (tan16 -tan9)f(tan^(-1)c)/(1+c^(2))` for some 'c' lying between tan9 and tan16
D. none of these

Answer

To solve the problem, we will analyze the given options based on the Mean Value Theorem for integrals, which states that if \( f(x) \) is continuous on the closed interval \([a, b]\), then there exists at least one \( c \) in \((a, b)\) such that:

\[ \int_a^b f(x) \, dx = (b - a) f(c) \]

### Step-by-Step Solution:

1. **Identify the Interval and Function**:
We have the interval \([9, 16]\) and the function \( f(x) \) which is continuous on this interval.

2. **Apply the Mean Value Theorem**:
According to the Mean Value Theorem:
\[ \int_9^{16} f(x) \, dx = (16 - 9) f(c) = 7 f(c) \]
for some \( c \) in the interval \((9, 16)\).

3. **Evaluate Option 1**:
The first option states:
\[ \int_9^{16} f(x) \, dx = 7 f(c) \]
Since we derived this from the Mean Value Theorem, **Option 1 is true**.

4. **Evaluate Option 2**:
The second option states:
\[ \int_9^{16} f(x) \, dx = \int_{\tan(9)}^{\tan(16)} f(\tan^{-1}(t)) \frac{1}{1 + t^2} \, dt \]
To verify this, we can use the substitution \( x = \tan^{-1}(t) \). The differential \( dx = \frac{1}{1 + t^2} dt \), and the limits change from \( x = 9 \) to \( x = 16 \) which correspond to \( t = \tan(9) \) to \( t = \tan(16) \). Thus, the equation holds true, making **Option 2 true**.

5. **Evaluate Option 3**:
The third option states:
\[ \int_9^{16} f(x) \, dx = (tan(16) - tan(9)) f(\tan^{-1}(c)) \frac{1}{1 + c^2} \]
Again, applying the Mean Value Theorem for the transformed integral:
\[ \int_{\tan(9)}^{\tan(16)} f(\tan^{-1}(t)) \frac{1}{1 + t^2} \, dt = (tan(16) - tan(9)) f(c') \]
for some \( c' \) in \((\tan(9), \tan(16))\). Since \( c' \) is a valid transformation of \( c \), **Option 3 is also true**.

6. **Conclusion**:
Since Options 1, 2, and 3 are all true, the answer is that all three options are correct.