Given a function 'g' continous everywhere such that `int _(0) ^(1) g (t ) dt =2 and g (1)=5.` If `f (x ) =1/2 int _(0) ^(x) (x-t) ^(2)g (t)dt,` then the vlaue of `f''(1)-f'(1)` is:

Options

A. 0
B. 1
C. 2
D. 10/7

Answer

To solve the problem, we need to find the value of \( f''(1) - f'(1) \) given the function:

\[ f(x) = \frac{1}{2} \int_0^x (x - t)^2 g(t) dt \]

where \( g(t) \) is continuous everywhere, \( \int_0^1 g(t) dt = 2 \), and \( g(1) = 5 \).

### Step 1: Determine the form of \( g(t) \)

Since we know that \( g(1) = 5 \) and \( \int_0^1 g(t) dt = 2 \), we can assume a simple polynomial form for \( g(t) \). Let's assume:

\[ g(t) = a t^b \]

From \( g(1) = 5 \), we have:

\[ g(1) = a \cdot 1^b = a = 5 \]

Thus, \( g(t) = 5 t^b \).

### Step 2: Use the integral condition

Now, we use the condition \( \int_0^1 g(t) dt = 2 \):

\[ \int_0^1 5 t^b dt = 2 \]

Calculating the integral:

\[ 5 \int_0^1 t^b dt = 5 \cdot \frac{1}{b + 1} = 2 \]

This gives us:

\[ \frac{5}{b + 1} = 2 \implies 5 = 2(b + 1) \implies 5 = 2b + 2 \implies 2b = 3 \implies b = \frac{3}{2} \]

Thus, we have:

\[ g(t) = 5 t^{3/2} \]

### Step 3: Substitute \( g(t) \) into \( f(x) \)

Now substituting \( g(t) \) into \( f(x) \):

\[ f(x) = \frac{1}{2} \int_0^x (x - t)^2 (5 t^{3/2}) dt = \frac{5}{2} \int_0^x (x - t)^2 t^{3/2} dt \]

### Step 4: Evaluate the integral

Using the binomial expansion:

\[ (x - t)^2 = x^2 - 2xt + t^2 \]

We can split the integral:

\[ f(x) = \frac{5}{2} \left( \int_0^x (x^2 t^{3/2}) dt - 2x \int_0^x (t^{5/2}) dt + \int_0^x (t^{7/2}) dt \right) \]

Calculating each integral:

1. \( \int_0^x t^{3/2} dt = \frac{2}{5} x^{5/2} \)
2. \( \int_0^x t^{5/2} dt = \frac{2}{7} x^{7/2} \)
3. \( \int_0^x t^{7/2} dt = \frac{2}{9} x^{9/2} \)

Substituting these back:

\[ f(x) = \frac{5}{2} \left( x^2 \cdot \frac{2}{5} x^{5/2} - 2x \cdot \frac{2}{7} x^{7/2} + \frac{2}{9} x^{9/2} \right) \]

Simplifying:

\[ = \frac{5}{2} \left( \frac{2}{5} x^{9/2} - \frac{4}{7} x^{9/2} + \frac{2}{9} x^{9/2} \right) \]

Combining the terms:

\[ = \frac{5}{2} x^{9/2} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) \]

### Step 5: Find \( f'(x) \) and \( f''(x) \)

Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{5}{2} \cdot \frac{9}{2} x^{7/2} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) \]

And differentiate again to find \( f''(x) \):

\[ f''(x) = \frac{5}{2} \cdot \frac{9}{2} \cdot \frac{7}{2} x^{5/2} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) \]

### Step 6: Evaluate at \( x = 1 \)

Finally, we evaluate \( f'(1) \) and \( f''(1) \):

1. Calculate \( f'(1) \)
2. Calculate \( f''(1) \)

Then compute:

\[ f''(1) - f'(1) \]

### Final Calculation

After evaluating, we find:

\[ f''(1) - f'(1) = \frac{10}{7} \]