The points of intersection of {F_1}(x) = int_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) First,evaluate the integrals for ${F_1}(x)$ and ${F_2}(x)$.${F_1}(x) = \int_2^x (2t - 5) dt = [t^2 - 5t]_2^x = (x^2 - 5x) - (2^2 - 5(2)) = x^2 - 5

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) First,evaluate the integrals for ${F_1}(x)$ and ${F_2}(x)$.${F_1}(x) = \int_2^x (2t - 5) dt = [t^2 - 5t]_2^x = (x^2 - 5x) - (2^2 - 5(2)) = x^2 - 5x + 6$.${F_2}(x) = \int_0^x 2t dt = [t^2]_0^x = x^2 - 0^2 = x^2$.To find the point of intersection,set ${F_1}(x) = {F_2}(x)$:$x^2 - 5x + 6 = x^2$.Subtracting $x^2$ from both sides,we get $-5x + 6 = 0$,which implies $5x = 6$,so $x = \frac{6}{5}$.Now,substitute $x = \frac{6}{5}$ into ${F_2}(x)$ to find the $y$-coordinate:$y = x^2 = \left( \frac{6}{5} \right)^2 = \frac{36}{25}$.Thus,the point of intersection is $\left( \frac{6}{5}, \frac{36}{25} \right)$.

The points of intersection of ${F_1}(x) = \int_2^x {(2t - 5)\,dt} $ and ${F_2}(x) = \int_0^x {2t\,dt} $ are

Similar Questions

If $\int \frac{dx}{1+\sin x} = \tan \left(\frac{x}{2}-\theta\right) + C$,then $\theta=$

If $\int \frac{e^x-1}{e^x+1} dx = f(x) + c$,then $f(x)$ is equal to

$\int \frac{dx}{\sqrt{7-6x-x^2}}$ is equal to

$\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx} = $

Find an anti-derivative (or integral) of the function $\sin 2x$ by the method of inspection.

Vedclass Test Series

Exam Paper Generator

Online Exam Module