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deff fn [24]
3 years ago
10

Evaluate using Definite integrals

Mathematics
1 answer:
swat323 years ago
4 0

Since [0,4]=[0,1]\cup(1,4], we can rewrite the integral as

\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt

Both integrals are quite immediate: you only need to use the power rule

\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}

to get

\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4

Now we only need to evaluate the antiderivatives:

\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15

So, the final answer is 15.

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8+30\2+4=27 put parentheses so it can equal 23 and 13
zhannawk [14.2K]

Answer:

correct anwer=27

Step-by-step explanation:

the anwer for this equation comes out to be 27. As simple as it becomes on applying paranthesis to this equation therefore we get

8+(30/2)+4=27

we need to apply the paranthesis using the method of BODMAS.

which favours division over addition.

therefore,correct anwer is 27 and not 13

6 0
3 years ago
Which of the following best defines an angle?
scoray [572]

Answer:

Third option

Step-by-step explanation:

The first definition is not as broad as it should be, therefore, while an angle does meet that criteria, it's not a general definition, hence, it is incorrect. The second one is incorrect as well because a parabola has a vertex, and it is certainly not an angle. This definition is not specific enough, so it will be eliminated. The last one is again, too specific. The third one is the best answer because it is broad enough while still being specific.

8 0
3 years ago
What number(s) have the absolute value of 7? Why? (Must explain to receive credit)
Liula [17]

Answer:

Why is | 7 | the absolute value of the number 7?

The absolute value of a number is how far away the number is from zero. The absolute value is ALWAYS positive.

What is the absolute value of -7?

The negative version of the number is also | 7 |

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Suppose your top drawer contains different colored socks: 14 are white, 4 are black, 12 are pink, and 8 are blue. All socks in t
Svetllana [295]

Answer:

8/38

Step-by-step explanation:

you calculate how many socks and then take how many the blue ones are

3 0
3 years ago
The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo
dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

\to f_X (x) \ \ xe^{-x} \ , \ x>0

For point a:

Moment generating function of X=?

Using formula:

\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx

M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx

integrating the values by parts:

u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0}  -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\  

        = \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\

Therefore, the moment value generating by the function is =\frac{1}{(t-1)^2}

In point b:

E(X^n)=?

Using formula: E(X^n)= M^{n}_{X}(0)

form point (a):

\to M_{X}(t)=\frac{1}{(t-1)^2}

Differentiating the value with respect of t

M'_{X}(t)=\frac{-2}{(t-1)^3}

when t=0

M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\

M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\

7 0
2 years ago
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