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x^{2}-x-6=0 -x+3\gt 2x+1 ; line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120)

Solve by substitution to find the intersection between the curves. Tap for more steps... Pull terms out from under the radical, assuming positive real numbers. Reorder 13 and - x2. The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region.

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