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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}{2...

The points of intersection ofF1(x)=∫2x(2t−5)dtF_{1}(x) = \int_{2}^{x}{(2t - 5)dt} and F2(x)=∫0x2tdt,F_{2}(x) = \int_{0}^{x}{2tdt,} are

A

(65,3625)\left( \frac{6}{5},\frac{36}{25} \right)

B

(23,49)\left( \frac{2}{3},\frac{4}{9} \right)

C

(13,19)\left( \frac{1}{3},\frac{1}{9} \right)

D

(15,125)\left( \frac{1}{5},\frac{1}{25} \right)

Answer

(65,3625)\left( \frac{6}{5},\frac{36}{25} \right)

Explanation

Solution

Let F1(x)=y1=∫2x(2t−5)dtF_{1}(x) = y_{1} = \int_{2}^{x}{(2t - 5)dt}and F2(x)=y2=∫0x2tdtF_{2}(x) = y_{2} = \int_{0}^{x}{2tdt}

Now point of intersection means those point at which y1=y2=y⇒y1=x2−5x+6y_{1} = y_{2} = y \Rightarrow y_{1} = x^{2} - 5x + 6 and y2=x2y_{2} = x^{2}.

On solving,

We get x2=x2−5x+6⇒x=65x^{2} = x^{2} - 5x + 6 \Rightarrow x = \frac{6}{5} and y=x2=3625y = x^{2} = \frac{36}{25}.

Thus point of intersection is(65,3625)\left( \frac{6}{5},\frac{36}{25} \right).