Categories > definite-integral > Summation of series by definite Integration > The points of intersection of ${{F}_{1}}(x)=\int_{2}^{x}{(2t-5)\,dt}$ and ${{F}_{2}}(x)=\int_{0}^{x}{2t\,dt,}$ are

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Question: The points of intersection of ${{F}_{1}}(x)=\int_{2}^{x}{(2t-5)\,dt}$ and ${{F}_{2}}(x)=\int_{0}^{x}{2t\,dt,}$ are


1) $\left( \frac{6}{5},\,\frac{36}{25} \right)$
2) $\left( \frac{2}{3},\,\frac{4}{9} \right)$
3) $\left( \frac{1}{3},\,\frac{1}{9} \right)$
4) $\left( \frac{1}{5},\,\frac{1}{25} \right)$
Solution: Explanation: No Explanation
Summation of series by definite Integration

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