§6.1 Using definite integrals to find area and length

Introduction

The Area Between Two Curves

Finding Area with Horizontal Slices

Finding the length of a curve

Find the area bounded between the graphs of $f(x) = (x-1)^2 + 1$ and $g(x) = x+2$.

Find the area of the region bounded by the parabola $x = y^2 - 1$ and the line $y = x-1$, shown at left in Figure 6.1.

To find where the graphs of $f(x) = 5-(x-1)^2$ and $g(x) = 4-x$ intersect, we set the formulas equal. The resulting intersection points are

The region bounded by $y = \sqrt{x}$ and $y = \frac{1}{4}x$ lies between the intersection points $x = 0$ and $x = 16$. Its area equals

The region bounded by $x = y^2$ and $x = 6-2y^2$ intersects at $y = \pm\sqrt{2}$. Using horizontal slices, the correct integrand is

The region bounded by the $y$-axis, $y = \cos(x)$, and $y = \sin(x)$ ends at their first positive intersection $x = \frac{\pi}{4}$. Compute its area, rounded to three decimals. The exact value is $\sqrt{2}-1$.

For $y = \sqrt{4-x^2}$ on $[-2,2]$, the arc length integral collapses because the curve is a semicircle of radius $2$. Compute the arc length, rounded to three decimals.

For most functions $f$, the arc length integral $\int_a^b\sqrt{1+(f'(x))^2}\,dx$ has no elementary antiderivative and must be approximated numerically.