Let p(x) be a function defined on R such that | vedclass

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

Question

(A) Given $p'(x) = p'(1 - x)$.Integrating both sides with respect to $x$,we get $p(x) = -p(1 - x) + C$.At $x = 0$,$p(0) = -p(1) + C$.Substituting the

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) Given $p'(x) = p'(1 - x)$.Integrating both sides with respect to $x$,we get $p(x) = -p(1 - x) + C$.At $x = 0$,$p(0) = -p(1) + C$.Substituting the given values $p(0) = 1$ and $p(1) = 41$,we have $1 = -41 + C$,which implies $C = 42$.Thus,$p(x) + p(1 - x) = 42$.Let $I = \int_{0}^{1} p(x) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have $I = \int_{0}^{1} p(1 - x) dx$.Adding the two expressions for $I$,we get $2I = \int_{0}^{1} (p(x) + p(1 - x)) dx$.Substituting $p(x) + p(1 - x) = 42$,we get $2I = \int_{0}^{1} 42 dx = 42[x]_{0}^{1} = 42$.Therefore,$I = 21$.

Let $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$ for all $x \in [0, 1]$,$p(0) = 1$,and $p(1) = 41$. Then $\int_{0}^{1} p(x) dx = $

Similar Questions

If $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$, then $g(x) =$

$\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx =$

$\int_0^{\pi} x f(\sin x) \, dx$ is equal to

If $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$, then,

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of the integral $\int_{-\pi / 2}^{\pi / 2} [[x] - \sin x] \, dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module