To find the area in the first quadrant bounded by the parabola \( y = x^2 + 1 \), the tangent to it at the point (2, 5), and the coordinate axes, we can follow these steps:
### Step 1: Find the equation of the tangent line at the point (2, 5).
1. **Differentiate the parabola**:
\[
y = x^2 + 1 \implies \frac{dy}{dx} = 2x
\]
At \( x = 2 \):
\[
\frac{dy}{dx} = 2 \cdot 2 = 4
\]
So, the slope of the tangent line at the point (2, 5) is 4.
2. **Use the point-slope form of the line**:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( m = 4 \), \( (x_1, y_1) = (2, 5) \):
\[
y - 5 = 4(x - 2)
\]
Simplifying gives:
\[
y = 4x - 3
\]
### Step 2: Find the points of intersection with the axes.
1. **Find the x-intercept** (set \( y = 0 \)):
\[
0 = 4x - 3 \implies 4x = 3 \implies x = \frac{3}{4}
\]
2. **Find the y-intercept** (set \( x = 0 \)):
\[
y = 4(0) - 3 = -3 \quad \text{(not relevant since we are in the first quadrant)}
\]
### Step 3: Set up the area calculation.
The area \( A \) in the first quadrant is bounded by the parabola from \( x = 0 \) to \( x = 2 \) and the tangent line from \( x = 0 \) to \( x = \frac{3}{4} \).
1. **Area under the parabola** from \( x = 0 \) to \( x = 2 \):
\[
A_1 = \int_0^2 (x^2 + 1) \, dx
\]
2. **Area under the tangent line** from \( x = 0 \) to \( x = \frac{3}{4} \):
\[
A_2 = \int_0^{\frac{3}{4}} (4x - 3) \, dx
\]
### Step 4: Calculate the areas.
1. **Calculate \( A_1 \)**:
\[
A_1 = \int_0^2 (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_0^2 = \left( \frac{2^3}{3} + 2 \right) - (0) = \left( \frac{8}{3} + 2 \right) = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}
\]
2. **Calculate \( A_2 \)**:
\[
A_2 = \int_0^{\frac{3}{4}} (4x - 3) \, dx = \left[ 2x^2 - 3x \right]_0^{\frac{3}{4}} = \left( 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) \right) - (0) = \left( 2 \cdot \frac{9}{16} - \frac{9}{4} \right)
\]
\[
= \left( \frac{18}{16} - \frac{36}{16} \right) = -\frac{18}{16} = -\frac{9}{8}
\]
### Step 5: Calculate the total area.
The area bounded by the parabola and the tangent line is:
\[
\text{Total Area} = A_1 - A_2 = \frac{14}{3} - \left(-\frac{9}{8}\right) = \frac{14}{3} + \frac{9}{8}
\]
To combine these fractions, find a common denominator (which is 24):
\[
\frac{14}{3} = \frac{112}{24}, \quad \frac{9}{8} = \frac{27}{24}
\]
So,
\[
\text{Total Area} = \frac{112}{24} + \frac{27}{24} = \frac{139}{24}
\]
### Final Answer:
The area in the first quadrant bounded by the parabola, the tangent line, and the coordinate axes is \( \frac{139}{24} \) square units.
