Applications of Integration

6.71 gallons

2.83 gallons

4.24 gallons

8.66 gallons

$\frac{d}{dt}(\sin(\pi t))\Big|_{0}^{\frac{a}{b}}$

$\int_{0}^{2\pi} \sin(\pi t), dt$

$(\sin(\pi \times \frac{a}{b}))-(\sin(\pi \times 0))$

$\int_{0}^{\frac{a}{b}} \sin(\pi t), dt$

75 gallons

100 gallons

150 gallons

200 gallons

INCORRECT ANSWER

$\int_7^{10}(6-0.8t) dt$

INCORRECT ANSWER

INCORRECT ANSWER

Has no effect on total costs since only marginal costs are increased but intervals stay constant.

Results in reduced total cost up to $q$ units due to decreased marginal costs.

Reduces slope increase per unit leading to flatter overall growth for $H(q)$.

Multiplies each incremental cost addition by $k$ resulting in steeper overall cost curve $H(q)$.

Volume of the solid object.

Mass of the solid object.

Length of the solid object.

Surface area of the solid object.

No lack of differentiability as long as $h(p)=15$ matches with $\int_{p}^{4} \frac{6}{t^{2}} dt$ when $p\geq4$.

At $p=0$ due to an undefined expression within the integrand.

At $p=4$ because $h(p)$ might not have a defined derivative there due to differing definitions.

For all $p<4$ since the integral involves division by zero within its limits.

The average rate of water flow during the first 10 minutes.

The total amount of water in the tank after some time has passed since it might not be empty initially.

The rate of water flow at exactly 10 minutes.

The total amount of water that flows into the tank during the first 10 minutes.

Graph $y=g'(x)$ traversed equal areas above below axis so their sum cancels out.

There was constant acceleration from $x$ Interval start until end point.

Function $g'$ reached its highest & lowest points equally often within $[5, 9]$.

Function $g$ had no net change within interval $[5, 9]$.

A triangle with base length r.

A semicircle with radius r.

A full circle with diameter r.

A rectangle with length r.