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Veseljchak [2.6K]
2 years ago
6

Find integra of xlnxdx

Mathematics
1 answer:
Mademuasel [1]2 years ago
5 0

Answer:

\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a <u>constant of integration</u>.

\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}

Increase the power by 1, then divide by the new power.

Given <u>indefinite integral</u>:

\displaystyle \int x \ln x \:\: \text{d}x

To integrate the given integral, use Integration by Parts:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}

\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2

Therefore:

\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}

Learn more about integration here:

brainly.com/question/27805589

brainly.com/question/27983581

brainly.com/question/27759474

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3 years ago
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masya89 [10]
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3 0
3 years ago
13 POINTS- please help me
allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures (6x+20)^{\circ} and (x+100)^{\circ} are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16

Then

(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}

17. Two parallel lines are cut by transversal. Angles with measures (2x+12)^{\circ} and (3x-22)^{\circ} are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34

Then

(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}

18. Two parallel lines are cut by transversal. Angles with measures (6x-7)^{\circ} and (5x+10)^{\circ} are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17

Then

(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}

19. The diagram shows two complementary angles with measures 2x^{\circ} and 56^{\circ}. The measures of complementary angles add up to 90^{\circ}, then

2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17

Hence,

2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}

Check:

34^{\circ}+56^{\circ}=90^{\circ}

20. Angles \angle 1 and \angle 2 are vertical angles. By vertical angles theorem, vertical angles are congruent, so

m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4

Hence,

m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}

21. \angle 5 and \angle 8 are supplementary. The measures of supplementary angles add up to 180^{\circ}, so

m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34

Therefore,

m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}

7 0
3 years ago
V-15=-27 what's they equation
andrezito [222]
V would be equal to -12
7 0
3 years ago
Read 2 more answers
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