What is the anti derivative of the square root of x dx?

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What is the anti derivative of the square root of x dx?

Calculus Integrals

2006-12-08 04:41:14 · 4 answers · asked by meg 1 in Science & Mathematics ➔ Mathematics

4 answers

∫x^1/2 dx = [x^(1/2+1)] / (1/2+1) + c

∫x^1/2 dx = 2[x^(1/2+2/2)] / (1/2+2/2) + c

∫x^1/2 dx = (2/3)*x^(3/2) + c

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2006-12-08 04:49:10 · answer #1 · answered by Luiz S 7 · 0⤊ 0⤋

The other answers are correct (make sure you put the +C)...it would help when you're finding antiderivatives if you changed the sqrt(x) to an x^1/2. Then, you treat it just like any other power of x...add one to the power, and divide by that number.

2006-12-08 12:49:53 · answer #2 · answered by Kim4 1 · 0⤊ 0⤋

(2/3)x^(3/2)

2006-12-08 12:44:56 · answer #3 · answered by n_m_young 4 · 0⤊ 0⤋

x^(3/2)/(3/2)+C
=(2/3)x^3/2+C

2006-12-08 12:43:22 · answer #4 · answered by raj 7 · 0⤊ 0⤋