1. Given curve and tangent

The parabola is $y=x^2+1.$

We need the tangent at the point $(2,5)$.

Since $\frac{dy}{dx}=2x,$ at $x=2$, the slope is $m=2(2)=4.$

So the tangent equation is $y-5=4(x-2),$ which simplifies to $y=4x-3.$

2. Find how the region is enclosed in the first quadrant

The boundaries are:

• the parabola $y=x^2+1$
• the tangent $y=4x-3$
• the coordinate axes $x=0$ and $y=0$

Let us locate key intersection points.

(i) Parabola with $y$-axis

At $x=0$, $y=0^2+1=1.$ So point is $(0,1)$.

(ii) Tangent with $x$-axis

Set $y=0$ in $y=4x-3$: $4x-3=0 \Rightarrow x=\frac34.$ So point is $\left(\frac34,0\right)$.

(iii) Tangent with parabola

Solve $x^2+1=4x-3.$ So $x^2-4x+4=0 \Rightarrow (x-2)^2=0.$ Thus they touch at $(2,5).$

Hence the enclosed region in the first quadrant is traced by:

• along the $y$-axis from $(0,0)$ to $(0,1)$,
• along the parabola from $(0,1)$ to $(2,5)$,
• along the tangent from $(2,5)$ to $\left(\frac34,0\right)$,
• along the $x$-axis from $\left(\frac34,0\right)$ to $(0,0)$.

3. Set up the area using vertical strips

For vertical strips, the lower boundary changes at $x=\frac34$:

• For $0\le x\le \frac34$, the strip runs from the parabola down to the $x$-axis. So area contribution is $\int_0^{3/4}(x^2+1)\,dx.$

• For $\frac34\le x\le 2$, the strip runs from the parabola down to the tangent. So area contribution is $\int_{3/4}^{2}\big[(x^2+1)-(4x-3)\big]dx.$

Therefore, $A=\int_0^{3/4}(x^2+1)dx+\int_{3/4}^{2}(x^2-4x+4)dx.$

Since $x^2-4x+4=(x-2)^2,$ we compute both parts.

4. Evaluate the integrals

First integral

$I_1=\int_0^{3/4}(x^2+1)dx=\left[\frac{x^3}{3}+x\right]_0^{3/4}.$

At $x=\frac34$, $\frac{(3/4)^3}{3}+\frac34=\frac{27}{64\cdot 3}+\frac34=\frac{9}{64}+\frac{48}{64}=\frac{57}{64}.$

So $I_1=\frac{57}{64}.$

Second integral

$I_2=\int_{3/4}^{2}(x-2)^2dx.$

Let $u=x-2$. Then an antiderivative is $\frac{(x-2)^3}{3}.$ Thus,

Now,