IF `int_0^a sqrtx dx=2a int_0^(pi//2) sin^3 x dx,` find `int_a^(a+1) x dx`.

Answer

To solve the problem step by step, we need to find the value of the integral \( \int_a^{a+1} x \, dx \) given that

\[ \int_0^a \sqrt{x} \, dx = 2a \int_0^{\frac{\pi}{2}} \sin^3 x \, dx. \]

### Step 1: Calculate \( \int_0^a \sqrt{x} \, dx \)

The integral can be computed as follows:

\[ \int_0^a \sqrt{x} \, dx = \int_0^a x^{\frac{1}{2}} \, dx = \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_0^a = \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_0^a = \frac{2}{3} a^{\frac{3}{2}}. \]

### Step 2: Calculate \( \int_0^{\frac{\pi}{2}} \sin^3 x \, dx \)

To compute \( \int_0^{\frac{\pi}{2}} \sin^3 x \, dx \), we can use the reduction formula or a known result. The integral can be evaluated as:

\[ \int_0^{\frac{\pi}{2}} \sin^3 x \, dx = \frac{3}{4}. \]

### Step 3: Set up the equation

Now substituting the results back into the original equation:

\[ \frac{2}{3} a^{\frac{3}{2}} = 2a \cdot \frac{3}{4}. \]

This simplifies to:

\[ \frac{2}{3} a^{\frac{3}{2}} = \frac{3}{2} a. \]

### Step 4: Solve for \( a \)

We can simplify this equation:

\[ \frac{2}{3} a^{\frac{3}{2}} - \frac{3}{2} a = 0. \]

Factoring out \( a \):

\[ a \left( \frac{2}{3} a^{\frac{1}{2}} - \frac{3}{2} \right) = 0. \]

This gives us two solutions:

1. \( a = 0 \) (not valid in this context)
2. \( \frac{2}{3} a^{\frac{1}{2}} = \frac{3}{2} \)

Solving for \( a \):

\[ 2a^{\frac{1}{2}} = \frac{9}{2} \implies a^{\frac{1}{2}} = \frac{9}{4} \implies a = \left(\frac{9}{4}\right)^2 = \frac{81}{16}. \]

### Step 5: Calculate \( \int_a^{a+1} x \, dx \)

Now we need to compute \( \int_a^{a+1} x \, dx \):

\[ \int_a^{a+1} x \, dx = \left[ \frac{x^2}{2} \right]_a^{a+1} = \frac{(a+1)^2}{2} - \frac{a^2}{2}. \]

Calculating this gives:

\[ = \frac{(a^2 + 2a + 1) - a^2}{2} = \frac{2a + 1}{2}. \]

Substituting \( a = \frac{81}{16} \):

\[ = \frac{2 \cdot \frac{81}{16} + 1}{2} = \frac{\frac{162}{16} + \frac{16}{16}}{2} = \frac{\frac{178}{16}}{2} = \frac{178}{32} = \frac{89}{16}. \]

### Final Answer

Thus, the value of \( \int_a^{a+1} x \, dx \) is

\[ \frac{89}{16}. \]

---