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\int \frac{1}{\sqrt{x}}\mathrm{d}x
Evaluate the indefinite integral first.
2\sqrt{x}
Rewrite \frac{1}{\sqrt{x}} as x^{-\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{-\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form.
2\times 5^{\frac{1}{2}}-2\times 1^{\frac{1}{2}}
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
2\sqrt{5}-2
Simplify.