§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

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 curves $y = (x-1)^2+1$ and $y = x+2$ intersect at $x = 0$ and $x = 3$, with the line on top between them. Compute the enclosed area as a decimal.

The region bounded by $y = 12-2x^2$ and $y = x^2-8$ runs between $x = \pm\sqrt{20/3}$. The correct area integral 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 $\sqrt{2}-1$.

The curves $y = x^3-x$ and $y = x^2$ intersect at $x = 0$ and $x = \frac{1\pm\sqrt{5}}{2}$, forming two separate regions. The total area between them equals

The region bounded by the parabola $x = y^2-1$ and the line $y = x-1$, rewritten as $x = y+1$, is best handled with horizontal slices. The correct integral is

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 $x = 1-y^2$ and $x = 2-2y^2$ intersects at $y = \pm1$, with the wider parabola on the right. Compute the area as a decimal, rounded to three places. The exact value is $\frac{4}{3}$.

The region bounded by the $x$-axis, $y = x^2$, and $y = 2-x$ is best sliced horizontally, writing $x = \sqrt{y}$ and $x = 2-y$. The area equals

The curves $x = y^2-2y$ and $y = x$, rewritten as $x = y$, intersect at $(0,0)$ and $(3,3)$. Using horizontal slices, the enclosed area equals

Using the arc length formula with $y' = 2x$, the length of $y = x^2$ from $x = -1$ to $x = 1$ is $\int_{-1}^{1}\sqrt{1+4x^2}\,dx$. Evaluate it with technology, rounded to three decimals.

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 $y = xe^{3x}$ on $[0,1]$, the product rule gives $y' = e^{3x}(3x+1)$. The correct arc length integral and its value are

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.

A particle moves at $7$ cm/sec along $y = 0.1x^2+1$ starting at $(0,1)$, so after $4$ seconds it has traveled $28$ cm of arc length. Its position satisfies