The area (in sq. units) 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 is:

We are asked to find the area that lies completely in the first quadrant and is enclosed by the parabola $$y = x^{2} + 1$$, its tangent at the point $$\bigl(2,\,5\bigr)$$ and the two coordinate axes.

To begin, we need the equation of the tangent. For a curve $$y=f(x)$$ the slope of the tangent at any point is given by the derivative $$\dfrac{dy}{dx}$$ evaluated at that point. Here $$f(x)=x^{2}+1,$$ so $$\dfrac{dy}{dx}=2x.$$

At the point $$x=2$$ (where the curve has the ordinate $$y=2^{2}+1=5$$) the slope is $$m = 2\!\times\!2 = 4.$$ Using the point-slope form of a straight line, $$y-y_{1}=m(x-x_{1}),$$ with $$\bigl(x_{1},y_{1}\bigr)=\bigl(2,5\bigr)$$ and $$m=4,$$ we obtain

$$y-5 = 4(x-2) \quad\Longrightarrow\quad y = 4x-3.$$

The two axes are $$x=0$$ (the y-axis) and $$y=0$$ (the x-axis). We now determine all the relevant points of intersection in the first quadrant (i.e. where $$x\ge 0$$ and $$y\ge 0$$).

1. Intersection of the parabola with the y-axis: put $$x=0$$ in $$y=x^{2}+1$$ to get $$y=1.$$ Hence the point is $$\bigl(0,1\bigr).$$

2. Intersection of the tangent with the x-axis: put $$y=0$$ in $$y=4x-3$$ to get $$0 = 4x-3 \;\;\Longrightarrow\;\; x = \dfrac34.$$ Hence the point is $$\Bigl(\dfrac34,\,0\Bigr).$$

3. The tangent and the parabola of course meet at the point of tangency $$\bigl(2,5\bigr).$$

Now we look at the shape that is enclosed. Starting from $$x=0$$ and moving rightwards:

• For $$0\le x\le \dfrac34$$ the tangent lies below the x-axis (its ordinate is negative there), so the lower boundary is the x-axis itself while the upper boundary is the parabola. Consequently, over this interval the vertical slice has height $$y_{\text{top}}-y_{\text{bottom}}=(x^{2}+1)-0.$$
• For $$\dfrac34\le x\le 2$$ the tangent has now risen above the x-axis. On this stretch the parabola is still above the tangent (they coincide only at $$x=2$$), so the slice height becomes $$y_{\text{top}}-y_{\text{bottom}}=(x^{2}+1)-(4x-3).$$

The total area is therefore the sum of two integrals.

First integral (between the parabola and the x-axis):

$$\begin{aligned} A_{1} &= \int_{0}^{3/4} \bigl(x^{2}+1\bigr)\,dx \\ &= \left[\frac{x^{3}}{3} + x\right]_{0}^{3/4} \\ &= \left(\frac{(3/4)^{3}}{3} + \frac34\right) - \left(0 + 0\right) \\ &= \left(\frac{27}{64}\cdot\frac13\right) + \frac34 \\ &= \frac{27}{192} + \frac{144}{192} \\ &= \frac{171}{192} \\ &= \frac{57}{64}. \end{aligned}$$

Second integral (between the parabola and the tangent):

First simplify the integrand:

$$\bigl(x^{2}+1\bigr)-(4x-3)=x^{2}-4x+4=(x-2)^{2}.$$

Hence

$$\begin{aligned} A_{2} &= \int_{3/4}^{2} (x-2)^{2}\,dx \\ &= \left[\frac{x^{3}}{3}-2x^{2}+4x\right]_{3/4}^{2}. \end{aligned}$$

Evaluating at the upper limit $$x=2$$ gives $$\frac{2^{3}}{3}-2(2)^{2}+4(2)=\frac83-8+8=\frac83.$$

Evaluating at the lower limit $$x=\frac34$$ gives

$$\begin{aligned} \frac{(3/4)^{3}}{3}-2\left(\frac34\right)^{2}+4\left(\frac34\right) &=\frac{27}{64}\cdot\frac13 - 2\cdot\frac{9}{16} + 3 \\ &=\frac{27}{192} - \frac{18}{16} + 3 \\ &=\frac{27}{192} - \frac{216}{192} + \frac{576}{192} \\ &=\frac{387}{192}. \end{aligned}$$

Therefore

$$A_{2}=\frac83 - \frac{387}{192} =\frac{512}{192} - \frac{387}{192} =\frac{125}{192}.$$

Total area:

$$\begin{aligned} A &= A_{1}+A_{2} \\ &= \frac{57}{64} + \frac{125}{192}. \end{aligned}$$

Convert $$\dfrac{57}{64}$$ to the common denominator $$192$$:

$$\frac{57}{64} = \frac{171}{192}.$$

Adding, we have

$$A = \frac{171}{192} + \frac{125}{192} = \frac{296}{192} = \frac{296\div 8}{192\div 8} = \frac{37}{24}.$$

Thus the required area in square units is $$\dfrac{37}{24}.$$

Hence, the correct answer is Option B.