(5)Value of `int_(0)^(2)3x^(2)dx+int_(0)^(pi//2)sinxdx`is

Answer

To solve the problem, we need to evaluate the following expression:

\[ \int_{0}^{2} 3x^{2} \, dx + \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 1: Evaluate the first integral \(\int_{0}^{2} 3x^{2} \, dx\)

1. **Factor out the constant**: The constant \(3\) can be taken outside the integral.
\[ \int_{0}^{2} 3x^{2} \, dx = 3 \int_{0}^{2} x^{2} \, dx \]

2. **Use the power rule for integration**: The integral of \(x^{n}\) is given by \(\frac{x^{n+1}}{n+1}\).
\[ \int x^{2} \, dx = \frac{x^{3}}{3} \]

3. **Apply the limits from \(0\) to \(2\)**:
\[ 3 \left[ \frac{x^{3}}{3} \right]_{0}^{2} = 3 \left( \frac{2^{3}}{3} - \frac{0^{3}}{3} \right) = 3 \left( \frac{8}{3} - 0 \right) = 8 \]

### Step 2: Evaluate the second integral \(\int_{0}^{\frac{\pi}{2}} \sin x \, dx\)

1. **Use the known integral of sine**:
\[ \int \sin x \, dx = -\cos x \]

2. **Apply the limits from \(0\) to \(\frac{\pi}{2}\)**:
\[ \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = -0 + 1 = 1 \]

### Step 3: Combine the results

Now, we combine the results of both integrals:
\[ \int_{0}^{2} 3x^{2} \, dx + \int_{0}^{\frac{\pi}{2}} \sin x \, dx = 8 + 1 = 9 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{9} \]