I need help with this and also round the Answer

Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

2 years ago

A machine making computer chips isn't working correctly, and 5% of the computer chips it makes are defective. If an inspector ch

Rainbow [258]

Answer:

0.25% probability that they are both defective

Step-by-step explanation:

For each computer chip, there are only two possible outcomes. Either they are defective, or they are not. The probability of a computer chip being defective is independent of other chips. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

5% of the computer chips it makes are defective.

This means that $p = 0.05$

If an inspector chooses two computer chips randomly (meaning they are independent from each other), what is the probability that they are both defective?

This is P(X = 2) when n = 2. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$

0.25% probability that they are both defective

4 0

2 years ago

Find the new amount of the original amount is 45 grams and the percent of decrease is 36%.

Semenov [28]

Answer:

28.8

Step-by-step explanation:

the original is 45=100%

it reduced by 36% meaning;100%-36%=64%

if45=100%

? =64%

45*64=2880/100=28.8

6 0

3 years ago

24 students in a class toook an algebra test if 18 students passed the test what percent passed

Sergio [31]

I’m pretty sure it’s 54%

6 0

2 years ago

What is 1 4/5 + 1 1/5??

Rudik [331]

$1\dfrac{4}{5}+1\dfrac{1}{5}=(1+1)+\left(\dfrac{4}{5}+\dfrac{1}{5}\right)=2+\dfrac{1+4}{5}=2+\dfrac{5}{5}=2+1=3$

8 0

2 years ago