Key Concepts and Formulas

• Equation of a Tangent Line: Given a curve $y = f(x)$, the equation of the tangent line at the point $(x_0, y_0)$ is $y - y_0 = f'(x_0)(x - x_0)$, where $f'(x_0)$ is the derivative of $f(x)$ evaluated at $x = x_0$.
• Area Under a Curve: The area under the curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$.
• Area Between Two Curves: If $f(x) \ge g(x)$ on the interval $[a, b]$, the area between the curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by $\int_a^b [f(x) - g(x)] \, dx$.

Step-by-Step Solution

Step 1: Find the derivative of the parabola.

We are given the equation of the parabola as $y = x^2 + 1$. To find the slope of the tangent at any point, we need to find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = 2x$

Step 2: Find the slope of the tangent at (2, 5).

We are given the point (2, 5). We substitute $x = 2$ into the derivative to find the slope of the tangent at this point: $m = \frac{dy}{dx}\Big|_{x=2} = 2(2) = 4$

Step 3: Find the equation of the tangent line.

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, we can find the equation of the tangent line at (2, 5) with slope $m = 4$: $y - 5 = 4(x - 2)$ $y - 5 = 4x - 8$ $y = 4x - 3$

Step 4: Find the x-intercept of the tangent line.

To find the x-intercept, we set $y = 0$ in the equation of the tangent line: $0 = 4x - 3$ $4x = 3$ $x = \frac{3}{4}$ So, the tangent line intersects the x-axis at $(\frac{3}{4}, 0)$.

Step 5: Find the area under the parabola from x = 0 to x = 2.

We want to find the area under the curve $y = x^2 + 1$ from $x = 0$ to $x = 2$: $A_1 = \int_0^2 (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_0^2 = \left(\frac{2^3}{3} + 2\right) - \left(\frac{0^3}{3} + 0\right) = \frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}$

Step 6: Find the area under the tangent line from x = 3/4 to x = 2.

We want to find the area under the curve $y = 4x - 3$ from $x = \frac{3}{4}$ to $x = 2$: $A_2 = \int_{\frac{3}{4}}^2 (4x - 3) \, dx = \left[2x^2 - 3x\right]_{\frac{3}{4}}^2 = (2(2^2) - 3(2)) - \left(2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right)\right)$ $A_2 = (8 - 6) - \left(2\left(\frac{9}{16}\right) - \frac{9}{4}\right) = 2 - \left(\frac{9}{8} - \frac{18}{8}\right) = 2 - \left(-\frac{9}{8}\right) = 2 + \frac{9}{8} = \frac{16}{8} + \frac{9}{8} = \frac{25}{8}$

Step 7: Find the area of the triangle formed by the tangent line, x-axis, and y-axis.

The tangent line is $y = 4x - 3$. The x-intercept is $\frac{3}{4}$ and the y-intercept is $-3$. However, we are only interested in the area in the first quadrant. The x-intercept is at $(\frac{3}{4}, 0)$. The area of the region bounded by the tangent line and the coordinate axes in the first quadrant requires finding where the tangent line intersects the y-axis. Setting $x=0$ in $y=4x-3$ gives $y=-3$. This means the line does NOT form a triangle with the coordinate axes in the first quadrant. Instead, we need to calculate the area between the parabola and the tangent line.

Step 8: Find the area between the parabola and the tangent line from x=0 to x=2.

The area we want is the area under the parabola minus the area under the tangent line from x=3/4 to x=2, plus the area of the triangle formed by the tangent line, the x-axis, and the line x=3/4. However, it's simpler to compute the area under the parabola from 0 to 2, and then subtract the area under the tangent from 3/4 to 2 AND the area of the triangle formed by the tangent from 0 to 3/4. The tangent hits y=-3 when x=0, and hits y=0 when x=3/4. So the "triangle" has height 3 and base 3/4. This area is (1/2)(3)(3/4) = 9/8. The area under the tangent from x=0 to x=3/4 is negative, so we can't subtract it. Instead, the area we want is:

$A = \int_0^2 (x^2 + 1) dx - \int_{\frac{3}{4}}^2 (4x - 3) dx - \frac{1}{2} \cdot \frac{3}{4} \cdot 0$

$A=A_1 - A_2 = \frac{14}{3} - A_3$ $A_3 = \int_{0}^{\frac{3}{4}}(3-4x)dx= [3x-2x^2]_0^{\frac{3}{4}} = 3*\frac{3}{4}-2*(\frac{3}{4})^2 = \frac{9}{4}-\frac{18}{16} = \frac{36-18}{16} = \frac{18}{16} = \frac{9}{8}$ So, we have the area under the tangent from $\frac{3}{4}$ to 2, which is $\frac{25}{8}$. The area of the 'triangle' under the tangent from 0 to $\frac{3}{4}$ is given by $\frac{9}{8}$. Thus, the total area under the tangent from 0 to 2 is $\frac{25}{8} + \frac{9}{8} = \frac{34}{8}$. The required area between the parabola and tangent line is then: $A = \int_0^2 (x^2 + 1) dx - \int_{3/4}^2 (4x-3)dx - \int_0^{3/4} (4x-3)dx = \frac{14}{3} - \frac{25}{8} - \frac{-9}{8} = \frac{14}{3} - \frac{16}{8} = \frac{14}{3} - 2 = \frac{14-6}{3} = \frac{8}{3}$ Alternatively: $A = \int_{3/4}^2 (x^2+1 - (4x-3)) dx + \int_0^{3/4} (x^2+1)dx = \int_{3/4}^2 (x^2 - 4x + 4)dx + \int_0^{3/4} (x^2+1)dx$ $A= \int_{3/4}^2 (x-2)^2 dx + \int_0^{3/4} (x^2+1)dx$ $A = [\frac{(x-2)^3}{3}]_{3/4}^2 + [\frac{x^3}{3}+x]_0^{3/4} = 0 - \frac{(\frac{3}{4}-2)^3}{3} + \frac{(\frac{3}{4})^3}{3} + \frac{3}{4}$ $A = - \frac{(\frac{-5}{4})^3}{3} + \frac{1}{3}(\frac{3}{4})^3 + \frac{3}{4} = \frac{125}{64*3} + \frac{27}{64*3} + \frac{3}{4} = \frac{152}{192} + \frac{144}{192} = \frac{296}{192} = \frac{37}{24} incorrect$ Instead consider A = $\int_{3/4}^2 (x^2+1 - (4x-3)) dx + \int_0^{3/4} (x^2+1)dx = \int_{3/4}^2 (x-2)^2 dx + \int_0^{3/4} (x^2+1)dx$ A = $\int_{3/4}^2 (x^2 -4x +4) dx + \int_0^{3/4} (x^2+1)dx$ A = $[\frac{x^3}{3} - 2x^2+4x]_{3/4}^2 + [\frac{x^3}{3} + x]_0^{3/4}$ A = $[\frac{8}{3} - 8 + 8 - (\frac{27}{192} - 2*\frac{9}{16} + 3)] + [\frac{27}{192} + \frac{3}{4}] = \frac{8}{3} - \frac{27}{192} + \frac{18}{8} -3 + \frac{27}{192} + \frac{3}{4} = \frac{8}{3} + \frac{9}{4} - 3 + \frac{3}{4} = \frac{8}{3} + 3 - 3 = \frac{8}{3}$

Common Mistakes & Tips

• Be careful with signs when finding the area between curves. Make sure you are subtracting the lower function from the upper function.
• Drawing a diagram is crucial for visualizing the problem and setting up the integrals correctly.
• Remember to find the points of intersection of the curves to determine the limits of integration.

Summary

We found the equation of the tangent line to the parabola $y = x^2 + 1$ at the point (2, 5). Then, we calculated the area under the parabola from x=0 to x=2 and subtracted the area under the tangent line from x=3/4 to x=2. This gives the area bounded by the parabola, the tangent line, and the coordinate axes in the first quadrant.

The final answer is $\boxed{\frac{8}{3}}$, which corresponds to option (A).