Find the average value of the function $f(t) = 10(e^{5t} - 1)$ over the interval $[0, 1]$.

Question

I have the following solution for my question but I am not sure if I have done the steps right:

Answer 1

You want to find $10\int (e^{5t} - 1) dt$. The anti-derivative is $10(\frac 15e^{5t} - t + c)$.

Taking bounds, we get: $10((\frac 15e^5 - 1) - (\frac 15e^0 - 0)) = 10(\frac 15(e^5 - 1) - 1) = 2(e^5 - 1) - 10$.

Since you're dividing by the interval $1-0 = 1$, that's also the required average. The approximate value is $284.83$

I spotted quite a few errors in your working. For one thing, you should be liberal in your use of brackets (parentheses) as it's not at all clear sometimes whether you mean ten times an entire expression or just one term. Near the start, you seem to have integrated the exponential term, but left the integral sign intact. Then you seem to have forgotten to integrate the $1$ (which should become $t$). You also seem to have added rather than subtracted (the values of the upper and lower bounds) when calculating the definite integral.

As I said, many errors. Please attempt it more systematically and carefully and see if you get the same answer I got.

Answer 2

Well, in general:

$$\hat{\text{y}}:=\frac{1}{\text{b}-\text{a}}\cdot\int_\text{a}^\text{b}\text{y}\left(x\right)\space\text{d}x\tag1$$

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So, when:

$$\text{y}\left(x\right):=\text{n}_1\cdot\left(\exp\left(\text{n}_2\cdot x+\text{n}_3\right)+\text{n}_4\right)\tag2$$

Then we know that:

$$\int_\text{a}^\text{b}\text{n}_1\cdot\left(\exp\left(\text{n}_2\cdot x+\text{n}_3\right)+\text{n}_4\right)\space\text{d}x=$$ $$\text{n}_1\cdot\left(\int_\text{a}^\text{b}\exp\left(\text{n}_2\cdot x+\text{n}_3\right)\space\text{d}x+\text{n}_4\cdot\int_\text{a}^\text{b}1\space\text{d}x\right)=$$ $$\text{n}_1\cdot\left(\int_\text{a}^\text{b}\exp\left(\text{n}_2\cdot x+\text{n}_3\right)\space\text{d}x+\text{n}_4\cdot\left[x\right]_\text{a}^\text{b}\right)=$$ $$\text{n}_1\cdot\left(\int_\text{a}^\text{b}\exp\left(\text{n}_2\cdot x+\text{n}_3\right)\space\text{d}x+\text{n}_4\cdot\left(\text{b}-\text{a}\right)\right)\tag3$$

Substitute $\text{u}:=\text{n}_2\cdot x+\text{n}_3$, so we get:

$$\int_\text{a}^\text{b}\exp\left(\text{n}_2\cdot x+\text{n}_3\right)\space\text{d}x=\frac{1}{\text{n}_2}\cdot\int_{\text{n}_2\cdot\text{a}+\text{n}_3}^{\text{n}_2\cdot\text{b}+\text{n}_3}\exp\left(\text{u}\right)\space\text{d}\text{u}=$$ $$\frac{1}{\text{n}_2}\cdot\left[\exp\left(\text{u}\right)\right]_{\text{n}_2\cdot\text{a}+\text{n}_3}^{\text{n}_2\cdot\text{b}+\text{n}_3}=\frac{\exp\left(\text{n}_2\cdot\text{b}+\text{n}_3\right)-\exp\left(\text{n}_2\cdot\text{a}+\text{n}_3\right)}{\text{n}_2}\tag4$$

So, we end up with:

$$\text{y}\left(x\right):=\text{n}_1\cdot\left(\frac{\exp\left(\text{n}_2\cdot\text{b}+\text{n}_3\right)-\exp\left(\text{n}_2\cdot\text{a}+\text{n}_3\right)}{\text{n}_2}+\text{n}_4\cdot\left(\text{b}-\text{a}\right)\right)\tag5$$

Now, in your problem $\text{n}_3=0$, $\text{n}_4=-1$, $\text{n}_1=10$ and $\text{n}_2=5$:

$$\text{y}\left(x\right):=10\cdot\left(\frac{\exp\left(5\cdot\text{b}\right)-\exp\left(5\cdot\text{a}\right)}{5}+\text{a}-\text{b}\right)\tag6$$

Answer 3

You made an error. Observe that

$ \int^1_0 (e^{5t} -1)dt = [\frac{e^{5t}}{5} - x]^1_0 = \frac{e^5 -1}{5} - 1 $

Observe that:

$[\frac{e^{5t}}{5} - x]^1_0 \neq [\frac{e^{5t}}{5} - 1]^1_0$