Question

Find the distance traveled in 15 seconds by an object moving with a velocity of v(t) = 20 + 7 cos t feet per second.

Answer 1

integrate
this is always positive, so the "object" is always moving forward
you need
\[\int_0^{15}(20+7\cos(t))dt\]

Answer 2

\[v(t)=\frac{ds}{dt}\]
Where s is displacement
So to get s integrate v(t) with respect to t.
\[\int\limits_{0}^{15}20+7cost=[20t+7sint](t=15)\]

Answer 3

not a bad integral since you can divide it up as
\[20\int_0^{15}1dt+7\int_0^{15}\cos(t)dt\]
first one is \(20\times 15=300\) and second one has anti derivative \(\sin(t)\) although you will need a calculator to compute \(\sin(15)\)

Answer 4

Could you please elaborate on what integrating is? I just opened up my calculus book for the first time today, and the chapter didn't say anything about integrals.

Answer 5

Derivatives are to do with the change of something (in this case, the change of position with respect to time (i.e. as time goes on)). The change of position is velocity, the change of velocity, acceleration.
Integrating is the reverse of differentiation (finding derivatives), and turns acceleration into velocity, and velocity into position.

Differentiation finds the gradient of a curve at a specific point (many functions are much harder than y=(gradient)x+c :think about x^2: the gradient is different for different points).

Integration finds the area under the curve \[\int\limits_{a}^{b}jkdlkdsjgh\] from x=a to x=b

An easy way to think about the symbols is that\[dx\]means 'a little bit of x'
and\[\int\limits_{}^{}\]means 'add up all the bits of'