Single Variable Calculus Questions and Answers

Start Your Free Trial

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

Expert Answers info

eNotes eNotes educator | Certified Educator

calendarEducator since 2007

write13,548 answers

starTop subjects are Math, Literature, and Science

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} $