Let f: R rightarrow R be defined as f(x)=e^{- | vedclass

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

Question

(B) Given $f(x) = e^{-x} \sin x$. Since $F(x) = \int_{0}^{x} f(t) dt$,by the Fundamental Theorem of Calculus,$F'(x) = f(x)$. We need to evaluate $I =

Vedclass · Accepted Answer

$[\frac{330}{360}, \frac{331}{360}]$

Vedclass · Answer

(B) Given $f(x) = e^{-x} \sin x$. Since $F(x) = \int_{0}^{x} f(t) dt$,by the Fundamental Theorem of Calculus,$F'(x) = f(x)$. We need to evaluate $I = \int_{0}^{1} (F'(x) + f(x)) e^{x} dx$. Substituting $F'(x) = f(x)$,we get $I = \int_{0}^{1} (f(x) + f(x)) e^{x} dx = \int_{0}^{1} 2f(x) e^{x} dx$. Substituting $f(x) = e^{-x} \sin x$,we get $I = 2 \int_{0}^{1} e^{-x} \sin x \cdot e^{x} dx = 2 \int_{0}^{1} \sin x dx$. Evaluating the integral,$I = 2 [-\cos x]_{0}^{1} = 2(1 - \cos 1)$. Using the Taylor series expansion for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$,we have $\cos 1 = 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \dots$. Thus,$I = 2(1 - (1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \dots)) = 2(\frac{1}{2} - \frac{1}{24} + \frac{1}{720} - \dots) = 1 - \frac{1}{12} + \frac{1}{360} - \dots$. $I = \frac{11}{12} + \frac{1}{360} - \dots = \frac{330}{360} + \frac{1}{360} - \dots = \frac{331}{360} - \dots$. Since the series is alternating and decreasing,the value lies in the interval $[\frac{330}{360}, \frac{331}{360}]$.

Let $f: R \rightarrow R$ be defined as $f(x)=e^{-x} \sin x$. If $F :[0,1] \rightarrow R$ is a differentiable function such that $F(x)=\int_{0}^{x} f(t) dt$,then the value of $\int_{0}^{1}(F'(x)+f(x)) e^{x} dx$ lies in the interval

Similar Questions

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$

Let $[x]$ denote the greatest integer less than or equal to $x$, then the value of the integral $\int_{-1}^{1}(|x|-2[x]) \, dx$ is equal to

The value of $x$ that satisfies the equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$ is

For a real number $x$,let $[x]$ denote the largest integer less than or equal to $x$ and $\{x\}=x-[x]$. The smallest possible integer value of $n$ for which $\int_1^n [x]\{x\} dx$ exceeds $2013$ is

The value of the integral $\int \limits_1^{\sqrt{2}+1} \left( \frac{x^2-1}{x^2+1} \right) \frac{1}{\sqrt{1+x^4}} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module