Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 48

Determine the area of the region bounded by the parabola $y = x^2$ and the tangent line to this parabola at $(1,1)$ and the $x$-axis.

We must first find the equation of the tangent line, so

$
\begin{equation}
\begin{aligned}
\text{if } y &= x^2, \text{ then}\\
\\
y' &= 2x
\end{aligned}
\end{equation}
$

Recall that $y ' = \text{ slope}$, so at point $(1,1)$
$y' = 2(1) = 2$
Using point slope form to determine the equation of the line...

$
\begin{equation}
\begin{aligned}
y - y_1 &= m(x - x_1) \\
\\
y - 1 &= 2(x- 1)\\
\\
y &= 2x - 1
\end{aligned}
\end{equation}
$


Lets graph these functions to evaluate the area easier



To evaluate the area without dividing into sub region, we must use horizontal strip for the area bounded by the curves including $x$-axis $(y = 0)$
So, $\displaystyle A = \int^{y_2}_{y_1} \left( x_{\text{right}} - x_{\text{left}} \right)dy$

$
\begin{equation}
\begin{aligned}
A &= \int^1_0 \left[ \frac{y+1}{2} - \sqrt{y} \right] dy\\
\\
A &= \int^1_0 \left[ \frac{y}{2} + \frac{1}{2} - (y)^{\frac{1}{2}} \right]dy\\
\\
A &= \left[ \frac{y^2}{2(2)} + \frac{1}{2}y - \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right]^1_0\\
\\
A &= 0.0833 \text{ square units}
\end{aligned}
\end{equation}
$