For an object whose velocity in ft/sec is given by v(t) = -2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

To find the distance traveled by an object on a certain time interval, we need to integrate the velocity function over that interval.

In this case, the velocity function is given as v(t) = -2t^2 + 4, where t represents time in seconds and v(t) represents the velocity at time t in feet per second.

The distance traveled by the object on the interval from t = 0 to t = 2 seconds can be found by evaluating the definite integral of the absolute value of the velocity function over that interval.

So, let's find the integral of |v(t)| from t = 0 to t = 2:

integral from 0 to 2 of |v(t)| dt = integral from 0 to 2 of |-2t^2 + 4| dt

To integrate the absolute value of a function, we need to split the interval into two parts, where the function is positive and where it is negative.

In this case, -2t^2 + 4 is positive on the interval from t = 0 to t = √(2), and it is negative on the interval from t = √(2) to t = 2.

So, we can rewrite the integral as follows:

integral from 0 to 2 of |-2t^2 + 4| dt = integral from 0 to √(2) of -(-2t^2 + 4) dt + integral from √(2) to 2 of -2t^2 + 4 dt

Simplifying and integrating each part separately:

= integral from 0 to √(2) of (2t^2 - 4) dt + integral from √(2) to 2 of (-2t^2 + 4) dt

= [2/3 * t^3 - 4t] evaluated from 0 to √(2) + [-2/3 * t^3 + 4t] evaluated from √(2) to 2

Evaluating the above expression:

= [2/3 * (√(2))^3 - 4√(2)] - [2/3 * 0^3 - 4 * 0] + [-2/3 * 2^3 + 4 * 2] - [-2/3 * (√(2))^3 + 4√(2)]

Simplifying further:

= [2/3 * 2√(2) - 4√(2)] + [-2/3 * 8 + 8] - [-2/3 * (√(2))^3 + 4√(2)]

= (4√(2)/3 - 4√(2)) + (-16/3 + 8) - ((-2/3)√(2) + 4√(2))

= (-8√(2)/3) + (8/3) - (8√(2)/3)

= -8√(2)/3 + 8/3 - 8√(2)/3

= -16√(2)/3 + 8/3

Therefore, the distance traveled by the object on the interval t = 0 to t = 2 seconds is -16√(2)/3 + 8/3 feet.

that would just be

∫[0,2] -2t^2+4 dt