Applications of Integration

1

1/3

1/6

1/2

From $\pi/6$ to $(5\pi)/6$

From $\pi/6$ to $\pi$

From $\pi/6$ to $(11\pi)/12$

From $\pi/6$ to $(7\pi)/6$

Calculating total length across intervals rather than areas.

Finding out which function is higher over each subinterval.

Determining which function has a greater numerical value at all points.

Identifying which function has a greater rate-of-change.

$\int_{0}^{3}( \sqrt{x}-\frac{1}{3}\sqrt{x})dx$

$\int_{-\infty}^{\infty}( \sqrt{x}-\frac{1}{3}x )dx$

$\int_{-9}^{-9}( \frac{1}{3}\sqrt{x}-\frac{9}{5})dx$

$\int_{0}^{9}( \sqrt{x}-\frac{1}{3}x )dx$

Multiply $(k-f(g(\frac{x}{c})))$ by $d$, integrate from $x=-d$ to $k$, and double result.

Integrate $(k-f(g(\frac{x}{c}))) dx$ from $x=-d$ to $\frac{d}{c}$ and double result since symmetric about y-axis.

Multiply $(k-f(g(\frac{x}{c})))$ by $c$ and double result and integrate $x=d$ to $k$.

Take definite integral of $(k-f(g(\frac{xd}{c})))$ from $x=-d$ to $d$, multiply by $c$, then double it.

2

1

$\frac{\pi}{2}$

4

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{12}$

$\frac{1}{20}$

$\frac{3}{2}$

$\frac{2}{3}$

$\frac{5}{3}$

$\frac{4}{3}$

Integrate $(4 - x^2) - (x^2)$ from $x=-1$ to $x=1$

Integrate $(4 - x^2) + (x^2)$ from $x=-1$ to $x=1$

Integrate $(4 - x^2) - (x^2)$ from $x=0$ to $x=1$

Integrate $(x^2) - (4-x^2)$ from $x=-1$ to $x=1$

Horizontal slicing causes undefined behaviors at intersection points whereas vertical slices maintain continuity across all sections being integrated.

Vertical slices result in integrals with respect to x, which avoids requiring inverse functions necessary for integrating with respect to y due to non-single-valued regions for $g(x)$.

Horizontal slicing leads directly to single variable integrals whereas vertical slicing results in multivariable calculus complications.

Vertical slicing does not apply here because both $\sqrt{x}$ and $\frac{x}{3}$ produce identical areas irrespective of slice orientation.