Let `f(x)=x^(3)+x^(2)+x+1,` then the area (in sq. units) bounded by `y=f(x), x=0, y=0 and x=1` is equal to

Options

A. `(25)/(3)`
B. `(25)/(12)`
C. `(12)/(5)`
D. `(5)/(3)`

Answer

To find the area bounded by the curve \( y = f(x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = 1 \), we will follow these steps:

1. **Define the Function**:
The given function is
\[ f(x) = x^3 + x^2 + x + 1. \]

2. **Identify the Area**:
The area \( A \) can be found using the definite integral of the function from \( x = 0 \) to \( x = 1 \):
\[ A = \int_{0}^{1} f(x) \, dx = \int_{0}^{1} (x^3 + x^2 + x + 1) \, dx. \]

3. **Integrate the Function**:
We will integrate each term of the function separately:
\[ \int (x^3 + x^2 + x + 1) \, dx = \int x^3 \, dx + \int x^2 \, dx + \int x \, dx + \int 1 \, dx. \]
The integrals are calculated as follows:
- \(\int x^3 \, dx = \frac{x^4}{4}\)
- \(\int x^2 \, dx = \frac{x^3}{3}\)
- \(\int x \, dx = \frac{x^2}{2}\)
- \(\int 1 \, dx = x\)

Therefore,
\[ \int (x^3 + x^2 + x + 1) \, dx = \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x. \]

4. **Evaluate the Definite Integral**:
Now we evaluate the integral from \( 0 \) to \( 1 \):
\[ A = \left[ \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x \right]_{0}^{1}. \]
Plugging in the limits:
\[ A = \left( \frac{1^4}{4} + \frac{1^3}{3} + \frac{1^2}{2} + 1 \right) - \left( \frac{0^4}{4} + \frac{0^3}{3} + \frac{0^2}{2} + 0 \right). \]
Simplifying this gives:
\[ A = \left( \frac{1}{4} + \frac{1}{3} + \frac{1}{2} + 1 \right). \]

5. **Finding a Common Denominator**:
To add these fractions, we need a common denominator. The least common multiple of \( 4, 3, 2, \) and \( 1 \) is \( 12 \):
- \(\frac{1}{4} = \frac{3}{12}\)
- \(\frac{1}{3} = \frac{4}{12}\)
- \(\frac{1}{2} = \frac{6}{12}\)
- \(1 = \frac{12}{12}\)

Now adding them together:
\[ A = \frac{3}{12} + \frac{4}{12} + \frac{6}{12} + \frac{12}{12} = \frac{25}{12}. \]

6. **Final Result**:
Therefore, the area bounded by the curve and the axes is
\[ \boxed{\frac{25}{12}} \text{ square units}. \]