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3 years ago

I really need these

Mathematics

1 answer:

Natalka [10]3 years ago

4 0

It’s a I hope you get a good gated

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How would I solve this

vova2212 [387]

$A=\frac{1}{2}\ln17 = 1.417$

Step-by-step explanation:

The area A under the curve can be written as

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}$

To evaluate the integral, let

$u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx$

so the integral becomes

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|$

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|$

Putting in the limits of integration, our area becomes

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2$

$\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)]$

$\;\;\;\;=\frac{1}{2}\ln17$

Note: $\ln 1 = 0$

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2 years ago

A rectangle in Quadrant II rotates 90° counterclockwise about the origin. In which Quadrant will the transformation lie?

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3 years ago

Find the inverse of the function x2 + y2 = 4 and the domain of the inverse for 0 ≤ x ≤ 2.

alexandr402 [8]

To get the inverse "relation" of an expression, we first off, do a quick switcharoo of the variables, and then solve for "y", so let's proceed,

$\bf x^2+y^2=4\qquad inverse\implies \boxed{y}^2+\boxed{x}^2=4
\\\\\\
y^2=4-x^2\implies y=\pm\sqrt{4-x^2}$

and yes, the domain for the range 0 ⩽ x ⩽ 2, let's get instead the "range" of the original function,

 $\bf x^2+y^2=4\implies 0^2+y^2=4\implies y=\pm\sqrt{4}\implies \boxed{y=\pm 2}
\\\\\\
x^2+y^2=4\implies 2^2+y^2=4\implies 4+y^2=4\implies \boxed{y=0}\\\\
-------------------------------\\\\
\stackrel{\textit{range of original}}{-2\le y \le 2}~~=~~\stackrel{\textit{domain of its inverse}}{-2\le x \le 2}$

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