If int_0^1 {{e^{{x^2}}}(x - alpha ),dx = 0,} | vedclass

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

Question

(C) Given $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0}$.This can be rewritten as $\int_0^1 {x{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} }$.For th

Vedclass · Accepted Answer

$0 < \alpha < 1$

Vedclass · Answer

(C) Given $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0}$.This can be rewritten as $\int_0^1 {x{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} }$.For the left side,let $u = x^2$,then $du = 2x \, dx$,so $x \, dx = \frac{1}{2} du$. When $x=0, u=0$ and when $x=1, u=1$.Thus,$\frac{1}{2} \int_0^1 {e^u du} = \alpha \int_0^1 {{e^{{x^2}}}dx}$.$\frac{1}{2} [e^u]_0^1 = \alpha \int_0^1 {{e^{{x^2}}}dx}$.$\frac{1}{2} (e - 1) = \alpha \int_0^1 {{e^{{x^2}}}dx}$.Therefore,$\alpha = \frac{e - 1}{2 \int_0^1 {{e^{{x^2}}}dx}}$.Since $e^x$ is an increasing function on $[0, 1]$,$1 \le e^{x^2} \le e$. Thus,$\int_0^1 1 \, dx Since $e \approx 2.718$,$\frac{e-1}{2} \approx 0.859$. Since $\int_0^1 e^{x^2} \, dx > 1$,$\alpha = \frac{0.859}{ ext{value} > 1} Also,since $e > 1$,$\alpha > 0$. Thus,$0 < \alpha < 1$.

If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

Similar Questions

The value of $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$ is

If $f(x) = \frac{e^x}{1 + e^x}$,$I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$,and $I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$,then the value of $\frac{I_2}{I_1}$ is

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

If $[x]$ is the greatest integer $\leq x$,then the value of the integral $\int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$ is

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module