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Arturiano [62]
3 years ago
6

An equation in the system is y = x + 1. What is the other equation in the system if the solution is only at (0, 1)?

Mathematics
1 answer:
Setler [38]3 years ago
6 0
<span>x + y = 0 hope this helps </span>
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3 years ago
\int (x+1)\sqrt(2x-1)dx
Nezavi [6.7K]

Answer:

\int (x+ 1) \sqrt{2x-1} dx =  \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C

Step-by-step explanation:

\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v =  \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}

\int (x+1)\sqrt(2x-1)dx\\\\   = uv - \int v du                              

= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  [ \ u = x + 1 => du = dx  \ ]    

= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\=  \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\

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Step-by-step explanation:

<em>It is assumed XYZ is a right triangle.</em>

  • y/10 = sin 32°
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7 0
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a cylinder has a volume of 108 cubic inches. the base has a radius of 3 inches and a height of 8 inches. what is the height of t
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Step-by-step explanation:

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