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3 years ago

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Daron purchased a cell phone case from the corner store. Daron paid with a $20 bill. The picture to the left shows his change. H

ow much was the cell phone case?

1 answer:

svlad2 [7]3 years ago

3 0

Do you have a picture of the question

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Given that f(x) = x2 + 1 and g(x) = -3, multiply the functions (gºf)(x).

stich3 [128]

Answer:

$(g\cdot f) = -3x^2-3$

Step-by-step explanation:

$f(x) = x^2 + 1$

$g(x) = -3$

The problem state to multiply the function $f$ and $g$ so the symbol seems to be incorrect.

We actually want to calculate

$g(x) f(x) =gf(x) = (-3) \cdot (x^2+1)= -3x^2-3$

Note

$g\circ f(x)=g(f(x))=g(x^2+1)=-3$

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2 years ago

Integrate sin^-1(x) dx<br><br> please explain how to do it aswell ...?

Lynna [10]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>

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3 years ago