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\int _{1}^{2}\sqrt{\pi }\mathrm{d}x
Cancel out x and x.
\int \sqrt{\pi }\mathrm{d}x
Evaluate the indefinite integral first.
\sqrt{\pi }x
Find the integral of \sqrt{\pi } using the table of common integrals rule \int a\mathrm{d}x=ax.
\pi ^{\frac{1}{2}}\times 2-\pi ^{\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.
\sqrt{\pi }
Simplify.