The area bounded by `f (x)=x^(2), 0 le x le 1, g(x)=-x+2,1 le x le 2` and x- axis is

Options

A. `(3)/(2)`
B. `(4)/(3)`
C. `(8)/(3)`
D. None of these

Answer

To find the area bounded by the curves \( f(x) = x^2 \) for \( 0 \leq x \leq 1 \) and \( g(x) = -x + 2 \) for \( 1 \leq x \leq 2 \) along with the x-axis, we will follow these steps:

### Step 1: Identify the area to be calculated
The area can be divided into two parts:
1. The area under the curve \( f(x) = x^2 \) from \( x = 0 \) to \( x = 1 \).
2. The area under the curve \( g(x) = -x + 2 \) from \( x = 1 \) to \( x = 2 \).

### Step 2: Set up the integrals for the areas
The area \( A \) can be expressed as:
\[ A = \int_0^1 f(x) \, dx + \int_1^2 g(x) \, dx \]

### Step 3: Calculate the first integral
For the first integral:
\[ A_1 = \int_0^1 x^2 \, dx \]
Using the power rule for integration:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
we have:
\[ A_1 = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Step 4: Calculate the second integral
For the second integral:
\[ A_2 = \int_1^2 (-x + 2) \, dx \]
This can be split into two parts:
\[ A_2 = \int_1^2 -x \, dx + \int_1^2 2 \, dx \]
Calculating each part:
1. For \( \int -x \, dx \):
\[ \int -x \, dx = -\frac{x^2}{2} + C \]
Thus,
\[ \left[ -\frac{x^2}{2} \right]_1^2 = -\frac{2^2}{2} + \frac{1^2}{2} = -2 + \frac{1}{2} = -\frac{4}{2} + \frac{1}{2} = -\frac{3}{2} \]

2. For \( \int 2 \, dx \):
\[ \int 2 \, dx = 2x + C \]
Thus,
\[ \left[ 2x \right]_1^2 = 2(2) - 2(1) = 4 - 2 = 2 \]

Combining both parts:
\[ A_2 = -\frac{3}{2} + 2 = -\frac{3}{2} + \frac{4}{2} = \frac{1}{2} \]

### Step 5: Combine the areas
Now, we can find the total area:
\[ A = A_1 + A_2 = \frac{1}{3} + \frac{1}{2} \]

### Step 6: Find a common denominator and simplify
The common denominator for \( \frac{1}{3} \) and \( \frac{1}{2} \) is 6:
\[ A = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]

### Final Answer
The area bounded by the curves and the x-axis is:
\[ \boxed{\frac{5}{6}} \text{ square units} \]