What is geometrical significance of
(i) `int_(0)^(pi) |cosx| dx`, (ii) ` int_(0)^((3pi)/(2)) cosx dx`

Answer

To solve the given question, we will analyze the two integrals separately and determine their geometrical significance.

### Part (i): \( \int_{0}^{\pi} |\cos x| \, dx \)

1. **Understanding the Function**: The function \( |\cos x| \) is the absolute value of \( \cos x \). The cosine function oscillates between -1 and 1. From \( x = 0 \) to \( x = \pi \), \( \cos x \) is positive from \( 0 \) to \( \frac{\pi}{2} \) and negative from \( \frac{\pi}{2} \) to \( \pi \). Therefore, the graph of \( |\cos x| \) will be the same as \( \cos x \) from \( 0 \) to \( \frac{\pi}{2} \) and will reflect the negative part from \( \frac{\pi}{2} \) to \( \pi \).

2. **Graphing the Function**:
- From \( 0 \) to \( \frac{\pi}{2} \), \( |\cos x| = \cos x \).
- From \( \frac{\pi}{2} \) to \( \pi \), \( |\cos x| = -\cos x \).

3. **Finding the Area**: The integral \( \int_{0}^{\pi} |\cos x| \, dx \) represents the total area under the curve \( y = |\cos x| \) from \( x = 0 \) to \( x = \pi \).
- The area from \( 0 \) to \( \frac{\pi}{2} \) is given by \( \int_{0}^{\frac{\pi}{2}} \cos x \, dx \).
- The area from \( \frac{\pi}{2} \) to \( \pi \) is given by \( \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \).

4. **Calculating the Areas**:
- Area from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]
- Area from \( \frac{\pi}{2} \) to \( \pi \):
\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = -[\sin x]_{\frac{\pi}{2}}^{\pi} = -(\sin(\pi) - \sin\left(\frac{\pi}{2}\right)) = - (0 - 1) = 1 \]

5. **Total Area**:
\[ \int_{0}^{\pi} |\cos x| \, dx = 1 + 1 = 2 \]

**Geometrical Significance**: The integral \( \int_{0}^{\pi} |\cos x| \, dx \) represents the total area bounded by the curve \( y = |\cos x| \) and the x-axis from \( x = 0 \) to \( x = \pi \), which is equal to 2 square units.

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### Part (ii): \( \int_{0}^{\frac{3\pi}{2}} \cos x \, dx \)

1. **Understanding the Function**: The function \( \cos x \) oscillates between -1 and 1. We need to analyze the behavior of \( \cos x \) from \( x = 0 \) to \( x = \frac{3\pi}{2} \).
- From \( 0 \) to \( \frac{\pi}{2} \), \( \cos x \) is positive.
- From \( \frac{\pi}{2} \) to \( \pi \), \( \cos x \) is zero at \( \frac{\pi}{2} \) and positive until \( \pi \).
- From \( \pi \) to \( \frac{3\pi}{2} \), \( \cos x \) is negative.

2. **Graphing the Function**:
- The area from \( 0 \) to \( \frac{\pi}{2} \) is positive.
- The area from \( \frac{\pi}{2} \) to \( \pi \) is also positive.
- The area from \( \pi \) to \( \frac{3\pi}{2} \) is negative.

3. **Finding the Areas**: The integral can be split into three parts:
\[ \int_{0}^{\frac{3\pi}{2}} \cos x \, dx = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx + \int_{\pi}^{\frac{3\pi}{2}} \cos x \, dx \]

4. **Calculating the Areas**:
- Area from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = 1 \]
- Area from \( \frac{\pi}{2} \) to \( \pi \):
\[ \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx = [\sin x]_{\frac{\pi}{2}}^{\pi} = 0 - 1 = -1 \]
- Area from \( \pi \) to \( \frac{3\pi}{2} \):
\[ \int_{\pi}^{\frac{3\pi}{2}} \cos x \, dx = [\sin x]_{\pi}^{\frac{3\pi}{2}} = -1 - 0 = -1 \]

5. **Total Area**:
\[ \int_{0}^{\frac{3\pi}{2}} \cos x \, dx = 1 - 1 - 1 = -1 \]

**Geometrical Significance**: The integral \( \int_{0}^{\frac{3\pi}{2}} \cos x \, dx \) represents the net area bounded by the curve \( y = \cos x \) and the x-axis from \( x = 0 \) to \( x = \frac{3\pi}{2} \). The positive area from \( 0 \) to \( \frac{\pi}{2} \) and \( \frac{\pi}{2} \) to \( \pi \) is outweighed by the negative area from \( \pi \) to \( \frac{3\pi}{2} \), resulting in a net area of -1 square units.

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