Integration and Accumulation of Change

Differential equation

Solution equation

Value theorem

It's differentiable everywhere even though it's not continuous at t=1.

It's continuous everywhere except at t=1 where it's not differentiable either.

It's neither continuous nor differentiable anywhere because it has a rational expression with variable in the denominator.

It’s discontinuous only where its derivative does not exist.

The average value of the function over a given interval

The accumulation of change in the function over a given interval

The maximum value of the function over a given interval

The rate of change of the function at a specific point

Doubling 'm' only affects cubic term magnitude leaving cosine term's derivative unchanged throughout domain.

Doubling 'm' multiplies original derivative by two and introduces additional oscillation within cosine term's derivative frequency.

When t = -4 seconds

When t = 4 seconds

When t = -2 seconds

Never; acceleration isn't zero at any time

The definite integral is equal to the average rate of change of the function over the interval.

The definite integral is equal to the maximum rate of change of the function over the interval.

The definite integral is equal to the derivative of the function evaluated at b minus the derivative of the function evaluated at a.

The definite integral is equal to the function evaluated at b minus the function evaluated at a.

Undefined because $f(x)$ has no minimum value as it decreases without bound.

$3$ because this is what you get when you integrate $f'(x)$ from 1 to infinity.

There is no minimum value since the function continues to increase as $x$ increases.

$0$ because the derivative approaches zero as $x$ increases.

The area under curve decreases as $(\ln(ax))^5$ becomes less steep with larger values of 'a'.

The area under the curve increases because $(\ln(ax))^5$ grows faster with larger values of 'a'.

The area remains constant as integral limits do not depend on variable 'a'.

There is no change to area until value 'k' also changes along with 'a'.

The definite integral depends on the value of the constant.

The definite integral is always positive.

The definite integral is always zero.

The definite integral is always negative.

$\int_{a}^{b} f(x) , dx$

$f(b) - f(a)$

$\frac{1}{b-a} \int_{a}^{b} f(x) , dx$

$f(b) + f(a)$