Let `f(x)` is a real valued function defined by `f(x)=x^(2)+x^(2) int_(-1)^(1) tf(t) dt+x^(3) int_(-1)^(1)f(t) dt`
then which of the following hold (s) good?

Options

A. ` int_(-1)^(1) tf(t) dt=(10)/(11)`
B. `f(1)+f(-1)=(30)/(11)`
C. `int_(-1)^(1)t f (t) dt gt int _(-1)^(1) f(t) dt`
D. `f(1)-f(-1)=(20)/(11)`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = x^2 + x^2 \int_{-1}^{1} t f(t) \, dt + x^3 \int_{-1}^{1} f(t) \, dt \]

Let's denote:

\[ a = \int_{-1}^{1} t f(t) \, dt \]
\[ b = \int_{-1}^{1} f(t) \, dt \]

Now, substituting these into the function \( f(x) \):

\[ f(x) = x^2 + ax^2 + bx^3 \]
\[ f(x) = (1 + a)x^2 + bx^3 \]

Next, we will find \( f(t) \) by substituting \( t \) for \( x \):

\[ f(t) = (1 + a)t^2 + bt^3 \]

Now, we need to compute \( a \) and \( b \):

### Step 1: Calculate \( a \)

Substituting \( f(t) \) into the expression for \( a \):

\[ a = \int_{-1}^{1} t f(t) \, dt = \int_{-1}^{1} t \left( (1 + a)t^2 + bt^3 \right) dt \]

This expands to:

\[ a = \int_{-1}^{1} \left( (1 + a)t^3 + bt^4 \right) dt \]

Now, we compute the integrals:

1. \(\int_{-1}^{1} t^3 \, dt = 0\) (since \( t^3 \) is an odd function)
2. \(\int_{-1}^{1} t^4 \, dt = \frac{2}{5}\)

Thus:

\[ a = 0 + b \cdot \frac{2}{5} \]
\[ a = \frac{2b}{5} \quad \text{(Equation 1)} \]

### Step 2: Calculate \( b \)

Now we compute \( b \):

\[ b = \int_{-1}^{1} f(t) \, dt = \int_{-1}^{1} \left( (1 + a)t^2 + bt^3 \right) dt \]

This expands to:

\[ b = (1 + a) \int_{-1}^{1} t^2 \, dt + b \int_{-1}^{1} t^3 \, dt \]

Again, we compute the integrals:

1. \(\int_{-1}^{1} t^2 \, dt = \frac{2}{3}\)
2. \(\int_{-1}^{1} t^3 \, dt = 0\)

Thus:

\[ b = (1 + a) \cdot \frac{2}{3} + 0 \]
\[ b = \frac{2}{3}(1 + a) \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:

1. \( a = \frac{2b}{5} \)
2. \( b = \frac{2}{3}(1 + a) \)

Substituting Equation 1 into Equation 2:

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

Multiply both sides by 15 to eliminate the denominators:

\[ 15b = 10(1 + \frac{2b}{5}) \]
\[ 15b = 10 + 6b \]
\[ 15b - 6b = 10 \]
\[ 9b = 10 \quad \Rightarrow \quad b = \frac{10}{9} \]

Now substituting \( b \) back into Equation 1 to find \( a \):

\[ a = \frac{2b}{5} = \frac{2 \cdot \frac{10}{9}}{5} = \frac{20}{45} = \frac{4}{9} \]

### Step 4: Check the options

Now we have:

- \( a = \frac{4}{9} \)
- \( b = \frac{10}{9} \)

We need to check the options based on these values.

1. **Option A**: \( a = b \) (False)
2. **Option B**: \( a < b \) (True, since \( \frac{4}{9} < \frac{10}{9} \))
3. **Option C**: \( a > b \) (False)
4. **Option D**: \( a + b = \frac{4}{9} + \frac{10}{9} = \frac{14}{9} \) (True)

Thus, the correct options are B and D.