IF `int_a^b x^1 dx=0 and int_a^b x^2 dx=2/3` then what are the values of a and b respectively.

Options

A. `-1.1`
B. 1,1
C. 0,0
D. 2,-2

Answer

To solve the problem, we need to evaluate the two given integrals and find the values of \( a \) and \( b \) based on the conditions provided.

### Step 1: Evaluate the first integral
The first integral is given as:

\[ \int_a^b x^1 \, dx = 0 \]

Calculating the integral:

\[ \int x^1 \, dx = \frac{x^2}{2} \]

Applying the limits from \( a \) to \( b \):

\[ \left[ \frac{b^2}{2} - \frac{a^2}{2} \right] = 0 \]

This simplifies to:

\[ \frac{b^2}{2} - \frac{a^2}{2} = 0 \]

Multiplying through by 2 gives:

\[ b^2 - a^2 = 0 \]

Factoring gives:

\[ (b - a)(b + a) = 0 \]

Thus, we have two cases:

1. \( b - a = 0 \) which implies \( b = a \)
2. \( b + a = 0 \) which implies \( b = -a \)

### Step 2: Evaluate the second integral
The second integral is given as:

\[ \int_a^b x^2 \, dx = \frac{2}{3} \]

Calculating the integral:

\[ \int x^2 \, dx = \frac{x^3}{3} \]

Applying the limits from \( a \) to \( b \):

\[ \left[ \frac{b^3}{3} - \frac{a^3}{3} \right] = \frac{2}{3} \]

This simplifies to:

\[ \frac{b^3}{3} - \frac{a^3}{3} = \frac{2}{3} \]

Multiplying through by 3 gives:

\[ b^3 - a^3 = 2 \]

### Step 3: Substitute values from the first integral
Now, we will consider the two cases from the first integral.

**Case 1:** \( b = a \)

Substituting \( b = a \) into \( b^3 - a^3 = 2 \):

\[ a^3 - a^3 = 2 \implies 0 = 2 \]

This is not possible.

**Case 2:** \( b = -a \)

Substituting \( b = -a \) into \( b^3 - a^3 = 2 \):

\[ (-a)^3 - a^3 = 2 \]

This simplifies to:

\[ -a^3 - a^3 = 2 \implies -2a^3 = 2 \]

Dividing both sides by -2 gives:

\[ a^3 = -1 \implies a = -1 \]

Now substituting \( a = -1 \) back to find \( b \):

\[ b = -(-1) = 1 \]

### Conclusion
Thus, the values of \( a \) and \( b \) are:

\[ \boxed{-1} \text{ and } \boxed{1} \]