Calculus I - Area Between Curves (Practice Problems)

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 6.2 : Area Between Curves

Determine the area below $f\left( x \right) = 3 + 2x - {x^2}$ and above the x-axis. Solution
Determine the area to the left of $g\left( y \right) = 3 - {y^2}$ and to the right of $x = - 1$ . Solution

For problems 3 – 11 determine the area of the region bounded by the given set of curves.

$y = {x^2} + 2$ , $y = \sin \left( x \right)$ , $x = - 1$ and $x = 2$ Solution
$\displaystyle y = \frac{8}{x}$ , $y = 2x$ and $x = 4$ Solution
$x = 3 + {y^2}$ , $x = 2 - {y^2}$ , $y = 1$ and $y = - 2$ Solution
$x = {y^2} - y - 6$ and $x = 2y + 4$ Solution
$y = x\sqrt {{x^2} + 1}$ , $y = {{\bf{e}}^{ - \,\,\frac{1}{2}x}}$ , $x = - 3$ and the y-axis. Solution
$y = 4x + 3$ , $y = 6 - x - 2{x^2}$ , $x = - 4$ and $x = 2$ Solution
$\displaystyle y = \frac{1}{{x + 2}}$ , $y = {\left( {x + 2} \right)^2}$ , $\displaystyle x = - \frac{3}{2}$ , $x = 1$ Solution
$x = {y^2} + 1$ , $x = 5$ , $y = - 3$ and $y = 3$ Solution
$x = {{\bf{e}}^{1 + 2y}}$ , $x = {{\bf{e}}^{1 - y}}$ , $y = - 2$ and $y = 1$ Solution