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4 years ago

10

Which is the binomial expansion (x+2)^4

1 answer:

prohojiy [21]4 years ago

8 0

x^4 +8x^3+24x^2+32x+16

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Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

3 years ago

C) does An altitude<br> Intersects the opposite side?

lora16 [44]

Answer:

An altitude of any triangle is a line from a vertex that intersects the opposite side at the right-angled vertex.

Therefore yes it does intersect the opposite side.

7 0

3 years ago

What number is: a composite, has 4 factors and some of it's multiples are 44 & 88?

Nadya [2.5K]

A delightful question.
I think I have it.

The number is 22 .

Its factors are 1, 2, 11, and 22 .

3 0

3 years ago

The subsets of the real numbers can be represented in a Venn diagram as shown. To which subset of real numbers does −2π belong?

Alecsey [184]

Answer:

D

Step-by-step explanation:

Whole numbers are the basic counting numbers: $0,\ 1,\ 2,\ 3,\ 4, \....$

Integers are whole numbers and negative numbers: $...,\ -2,\ -1,\ 0,\ 1,\ 2,\ ...$

Rational numbers are all fractions $\frac{m}{n},$ where m is an integer and n is a natural number.

Irrational numbers are all real numbers that are not rational.

Number $\pi$ is irrational number (it is not counting number, so it is not whole number, not integer number), it cannot be represented as a fraction, so it is not rational number, thus this number is irrational number.

The product of integer number -2 by irrational number $\pi$ is irrational number $-2\pi$

Correct option - option D

7 0

3 years ago

Read 2 more answers

Please i need the answer

Harrizon [31]

Answer: b
sorry cant give explanation

7 0

3 years ago