Let f(x) and g(x) be two functions satisfying | vedclass

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

Question

(D) Given $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) = -g(x)$.Substituting $g(4-x) = -g(x)$ into the first equation,we get $f(x^{2}) - g(x) = 4x^{3}$,so

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(D) Given $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) = -g(x)$.Substituting $g(4-x) = -g(x)$ into the first equation,we get $f(x^{2}) - g(x) = 4x^{3}$,so $f(x^{2}) = 4x^{3} + g(x)$.We want to evaluate $I = \int_{-4}^{4} f(x) dx$. Let $x = t^{2}$,then $dx = 2t dt$. However,since $f(x)$ is defined via $f(x^{2})$,we consider the integral $\int_{-4}^{4} f(x^{2}) dx$.Since $f(x^{2}) = 4x^{3} + g(x)$,we have $\int_{-4}^{4} f(x^{2}) dx = \int_{-4}^{4} (4x^{3} + g(x)) dx$.$= \int_{-4}^{4} 4x^{3} dx + \int_{-4}^{4} g(x) dx$.Since $4x^{3}$ is an odd function,$\int_{-4}^{4} 4x^{3} dx = 0$.Given $g(4-x) = -g(x)$,let $u = 4-x$,then $du = -dx$. The integral $\int_{-4}^{4} g(x) dx$ becomes $\int_{8}^{0} g(4-u) (-du) = \int_{0}^{8} g(4-u) du = \int_{0}^{8} -g(u) du$. This implies the integral of $g(x)$ over symmetric intervals is $0$.Thus,the integral evaluates to $0$.

Let $f(x)$ and $g(x)$ be two functions satisfying $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) + g(x) = 0$. Then the value of $\int_{-4}^{4} f(x) dx$ is

Similar Questions

Let $[t]$ denote the greatest integer less than or equal to $t$. Then the value of the integral $\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x$ is equal to

$\int_{-\pi / 15}^{\pi / 15} \frac{\cos 5 x}{1+e^{5 x}} d x=$

Let $T > 0$ be a fixed number. $f: R \rightarrow R$ is a continuous function such that $f(x+T) = f(x)$ for all $x \in R$. If $I = \int_0^T f(x) dx$,then find the value of $\int_0^{5T} f(2x) dx$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx =$

$\int_{-1}^{3} \left[ \tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \tan^{-1} \left( \frac{x^{2}+1}{x} \right) \right] dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module