Please help!

Mathematics

1 answer:

Murrr4er [49]1 year ago

3 0

The integral which expresses the area of the region bounded by y = ¹/₂x - 2 and y = ∛x is; ∫⁸₀((x^(1/3)) - ¹/₂x + 2

<h3>How to integrate between two curves?</h3>

We are given the functions as;

y = ¹/₂x - 2

y = ∛x

Now, from the graph the coordinate of the point where both curves intersect is at; (8, 2)

Thus, the x-coordinate here is x = 8

∛x - (¹/₂x - 2)

∛x - ¹/₂x + 2

Similarly, another point of intersection of both curves is at the coordinate (0, -2). Thus, the x-coordinate here is; x = 0

Thus, the integral which expresses the area of the region bounded by y = ¹/₂x - 2 and y = ∛x is; ∫⁸₀((x^(1/3)) - ¹/₂x + 2

Read more about Integration at; brainly.com/question/16415788

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Answer: x = 0

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$3^{2x}+3^{x+1}-4=0\\\\3^{2x}+3^x\cdot3^1-4=0\\\\\text{Let u = }3^x\\u^2+3u-4=0\\\\(u+4)(u-1)=0\\\\u+4=0\quad and\quad u-1=0\\u=-4\qquad and\quad u=1\\\\\text{Substitute u with }3^x\\3^x=-4\qquad and\quad 3^x=1\\\text{not possible}\ and\quad 3^x=3^0\\.\qquad \qquad \qquad \qquad \boxed{x=0}$

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Answer: No Solution

Step-by-step explanation:

$4^x+6\cdot 2^x+8=0\\\\2\cdot 2^x+6\cdot 2^x+8=0\\\\\text{Let u = }2^x\\2u+6u+8=0\\8u+8=0\\8u=-8\\u=-1\\\\\text{Substitute u with }2^x\\2^x=-1\\\text{not possible}$