`int_(0)^(pi) k(pix-x^(2))^(100)sin2x" dx"` is equal to

Options

A. `pi^(100)`
B. `(1)/(2)(pi^(100)-pi^(97))`
C. `(1)/(2)(pi^(100)+pi^(97))`
D. 0

Answer

To solve the integral \( I = \int_{0}^{\pi} k(\pi x - x^2)^{100} \sin(2x) \, dx \), we can use a property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

### Step 1: Apply the property of definite integrals

Let’s denote our integral as:

\[ I = \int_{0}^{\pi} k(\pi x - x^2)^{100} \sin(2x) \, dx \]

Now, applying the property, we have:

\[ I = \int_{0}^{\pi} k(\pi(\pi - x) - (\pi - x)^2)^{100} \sin(2(\pi - x)) \, dx \]

### Step 2: Simplify the expression

Now, we simplify the expression inside the integral:

1. Calculate \( \pi(\pi - x) - (\pi - x)^2 \):
\[ = \pi^2 - \pi x - (\pi^2 - 2\pi x + x^2) \]
\[ = \pi^2 - \pi x - \pi^2 + 2\pi x - x^2 \]
\[ = \pi x - x^2 \]

2. For \( \sin(2(\pi - x)) \):
\[ = \sin(2\pi - 2x) = -\sin(2x) \]

### Step 3: Substitute back into the integral

Now substituting these back into the integral gives us:

\[ I = \int_{0}^{\pi} k(\pi x - x^2)^{100} (-\sin(2x)) \, dx \]
\[ = -\int_{0}^{\pi} k(\pi x - x^2)^{100} \sin(2x) \, dx \]
\[ = -I \]

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

From the equation \( I = -I \), we can add \( I \) to both sides:

\[ 2I = 0 \]

Thus, we find:

\[ I = 0 \]

### Final Answer

The value of the integral is:

\[ \int_{0}^{\pi} k(\pi x - x^2)^{100} \sin(2x) \, dx = 0 \]