Question

Find the average value of the function below over the interval [0 , 16].

g(t) = 1 + t

Answer 1

The average value of \(f(x)\) over some interval \([a,b]\) is defined by \(\large f_{\text{ave}} = \displaystyle \frac{1}{b-a}\int_a^b f(x)\,dx\). In your problem, you want to compute \(\large f_{\text{ave}} = \displaystyle \frac{1}{16}\int_0^{16} 1+t\,dt\).

Can you take things from here? :-)

Answer 2

I'm still very weak in this area. How do I go about this?

Answer 3

I think i understand the b-a part

Answer 4

Do you know how to integrate \(\large\displaystyle \int_0^{16} 1+t\,dt\)?

Answer 5

not really. is it the anti derivative?

Answer 6

That would be correct; do you know the power rule for integration?

Answer 7

no, don't think so. we just started this recently and im confused

Answer 8

Okay. The power rule for integration is the following: \[\large \int x^n\,dx = \frac{x^{n+1}}{n+1}+C,\quad n\neq -1.\]So to integrate \(\large \displaystyle \int 1+t\,dt\), you can treat it as two separate integrals (by linearity of integration) and instead evaluate \(\large\displaystyle \int 1\,dt + \int t\,dt\). Now, treat \(\large 1=t^0\); we now have \(\large\displaystyle\int t^0\,dt + \int t\,dt= \frac{t^{0+1}}{0+1} + \frac{t^{1+1}}{1+1}+C = t + \frac{1}{2}t^2+C\).

However, we're evaluating the definite integral \(\large\displaystyle \int_0^{16} 1+t\,dt\), so once we have the antiderivative (i.e. \(\large t+\frac{1}{2}t^2\)), we then use the fundamental theorem of calculus to get the answer we want. In particular, \(\large\displaystyle \int_a^bf(t)\,dt = F(b)-F(a)\) where \(\large F(t)\) is our antiderivative. Thus, we see in our problem that\[\begin{aligned}\int_0^{16}1+t\,dt &= \left.\left(t+\frac{1}{2}t^2\right)\right|_0^{16}\\ &= \left((16)+\frac{1}{2}(16)^2\right) - \left((0) + \frac{1}{2}(0)^2\right)\\ &= 144\end{aligned}\]Therefore, \[\large f_{\text{ave}} = \frac{1}{16}\int_0^{16} 1+t\,dt = \frac{1}{16}\cdot 144 = 9.\]Does this make sense? :-)

Answer 9

If you haven't seen the Fundamental theorem of calculus yet, there is an alternative way of doing this, so let me know if that's the case. :-)

Answer 10

Thanks so much!

Answer 11

@ChristopherToni

Answer 12

i have another similar question, but I tried following your steps and didn't get it right

Answer 13

Find the average value of the function below over the interval [0 , 3].
f(x) = 2x + 8

Answer 14

the anti derivative is x^2 + x?

Answer 15

then i did 3^(2) + 3 - 0^2 + 0

Answer 16

times 1/3

Answer 17

The antiderivative is \(x^2+\color{red}8x\). :-)

In general whenever you integrate a constant (i.e. a real number all by itself), the antiderivative is the constant times x; i.e. the antiderivative of \(c\) is \(cx\).

Does that clarify things?

Answer 18

im still not sure how you got that

Answer 19

oh, u used the constant?

Answer 20

the answer would be 11?

Answer 21

@ChristopherToni

Answer 22

Note that \[\begin{aligned}\int 2x+8\,dx &= \int 2x\,dx + \int 8\,dx\\ &= 2\cdot\frac{x^2}{2} + 8 \cdot\frac{x^1}{1} + C\\ &= x^2+8x+C\end{aligned}\]However, since you're computing a definite integral, we don't include the +C when we find the antiderivative.

Then it follows that \[\begin{aligned} f_{\text{ave}} &= \frac{1}{3}\int_0^3 2x+8\,dx\\ & = \frac{1}{3}\left.\left(x^2+8x\right)\right|_0^3\\ &= \frac{1}{3}\left[ (3^2+8(3)) - (0^2-8(0))\right]\\ &=\frac{1}{3} \cdot 33 = 11\end{aligned} \]Does this clarify things? :-)

Answer 23

Oh, ok I think I kind of get it now. Thanks!