Area of the region bounded by curves `y=x^(2)+2,y=x,x=0 and x=1 is 7/16`.True or False

Answer

To determine whether the statement "The area of the region bounded by the curves \( y = x^2 + 2 \), \( y = x \), \( x = 0 \), and \( x = 1 \) is \( \frac{7}{16} \)" is true or false, we will calculate the area step by step.

### Step 1: Identify the curves and the area to be calculated
We have the curves:
- \( y = x^2 + 2 \) (a parabola opening upwards)
- \( y = x \) (a straight line)
- Vertical lines \( x = 0 \) and \( x = 1 \)

We need to find the area between these curves from \( x = 0 \) to \( x = 1 \).

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine where the curves intersect. Set \( x^2 + 2 = x \):
\[ x^2 - x + 2 = 0 \]
Calculating the discriminant:
\[ D = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 \]
Since the discriminant is negative, the curves do not intersect.

### Step 3: Determine which curve is on top
For \( x \) in the interval \( [0, 1] \):
- At \( x = 0 \): \( y = 0^2 + 2 = 2 \) and \( y = 0 \) (the line)
- At \( x = 1 \): \( y = 1^2 + 2 = 3 \) and \( y = 1 \)

Thus, \( y = x^2 + 2 \) is above \( y = x \) in the interval \( [0, 1] \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} ((x^2 + 2) - x) \, dx \]

### Step 5: Calculate the integral
Now, simplify the integrand:
\[ A = \int_{0}^{1} (x^2 + 2 - x) \, dx = \int_{0}^{1} (x^2 - x + 2) \, dx \]

Now, compute the integral:
\[ A = \int_{0}^{1} (x^2 - x + 2) \, dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_{0}^{1} \]
Calculating the upper limit:
\[ A = \left( \frac{1^3}{3} - \frac{1^2}{2} + 2 \cdot 1 \right) - \left( \frac{0^3}{3} - \frac{0^2}{2} + 2 \cdot 0 \right) \]
\[ = \left( \frac{1}{3} - \frac{1}{2} + 2 \right) \]
Finding a common denominator (6):
\[ = \left( \frac{2}{6} - \frac{3}{6} + \frac{12}{6} \right) = \frac{2 - 3 + 12}{6} = \frac{11}{6} \]

### Step 6: Conclusion
The area \( A = \frac{11}{6} \) square units, which is not equal to \( \frac{7}{16} \).

Thus, the statement is **False**.