Question

The value of \(\int\limits_{0}^{\frac{\pi}{2}}f(x)g(x)dx-\int\limits_{0}^{\frac{\pi}{2}}g(x)dx\) is ______.

Answer 1

Correct Answer: 0

Integration of \( f(x)g(x) \) 

Let:

\[ I = \int_0^{\frac{\pi}{2}} f(x) g(x) \, dx \]

Substituting \( f(x) = \sin^2(x) \) and \( g(x) = \pi x^2 - x^2 \), we get:

\[ I = \int_0^{\frac{\pi}{2}} \sin^2(x) \left( \pi x^2 - x^2 \right) \, dx \]

Using the trigonometric identity \( \sin^2(x) = 1 - \cos^2(x) \):

\[ I = \int_0^{\frac{\pi}{2}} \cos^2(x) \left( \pi x^2 - x^2 \right) \, dx \]

Combining integrals:

\[ 2I = \int_0^{\frac{\pi}{2}} \left( \sin^2(x) + \cos^2(x) \right) \left( \pi x^2 - x^2 \right) \, dx \]

This simplifies to:

\[ 2I = \int_0^{\frac{\pi}{2}} g(x) \, dx \]

Thus, we have:

\[ 2 \int_0^{\frac{\pi}{2}} f(x) g(x) \, dx - \int_0^{\frac{\pi}{2}} g(x) \, dx = 0 \]

Answer 2

Calculation of \( I \) 

We are given:

\[ f(x) = \sin^2(x), \quad g(x) = \pi x^2 - x^2 \]

And:

\[ I = \int_0^{\frac{\pi}{2}} \sin^2(x) \left( \pi x^2 - x^2 \right) \, dx \]

Step 1: Apply Trigonometric Identity

Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), we write:

\[ I = \int_0^{\frac{\pi}{2}} \cos^2(x) \left( \pi x^2 - x^2 \right) \, dx \]

Step 2: Combine the Two Forms of \( I \)

Adding the two integrals for \( \sin^2(x) \) and \( \cos^2(x) \), we get:

\[ 2I = \int_0^{\frac{\pi}{2}} \left( \sin^2(x) + \cos^2(x) \right) \left( \pi x^2 - x^2 \right) \, dx \]

Since \( \sin^2(x) + \cos^2(x) = 1 \), the integral simplifies to:

\[ 2I = \int_0^{\frac{\pi}{2}} \pi x^2 - x^2 \, dx \]

Step 3: Evaluate the Integral

The integral \( \int_0^{\frac{\pi}{2}} \pi x^2 - x^2 \, dx \) is known to equal \( \frac{\pi^3}{32} \). Substituting this result:

\[ 2I = 16 \cdot \frac{\pi^3}{32} \]

Simplify:

\[ 2I = \frac{16}{32} = \frac{1}{2} \]

Step 4: Solve for \( I \)

Divide both sides by 2:

\[ I = \frac{1}{4} \]

Final Answer:

• \( I = 0.25 \)