If I = int_{0}^{2} e^{x^{4}}(x - alpha) dx = | vedclass

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

Question

(A) Given the integral equation: $\int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$. This can be rewritten as: $\int_{0}^{2} x e^{x^{4}} dx = \alpha \int_{0}

Vedclass · Accepted Answer

$(0, 2)$

Vedclass · Answer

(A) Given the integral equation: $\int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$. This can be rewritten as: $\int_{0}^{2} x e^{x^{4}} dx = \alpha \int_{0}^{2} e^{x^{4}} dx$. Thus, $\alpha = \frac{\int_{0}^{2} x e^{x^{4}} dx}{\int_{0}^{2} e^{x^{4}} dx}$. Let $f(x) = e^{x^{4}}$. Since $f(x) > 0$ for all $x \in [0, 2]$, the expression for $\alpha$ represents a weighted average of the values of $x$ in the interval $[0, 2]$ with respect to the weight function $f(x)$. Since $x$ ranges from $0$ to $2$, the weighted average $\alpha$ must also lie strictly between the minimum and maximum values of $x$ in the interval $[0, 2]$. Therefore, $0 < \alpha < 2$, which means $\alpha \in (0, 2)$.

If $I = \int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$, then $\alpha$ lies in the interval

Similar Questions

$\int_{0}^{2\pi} |\sin x| \, dx = $

The value of $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$ is

The integral $80 \int_0^{\frac{\pi}{4}} \left( \frac{\sin \theta + \cos \theta}{9 + 16 \sin 2 \theta} \right) d \theta$ is equal to :

Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$

Let $\int_0^1 f(x) \, dx = 1$,$\int_0^1 x f(x) \, dx = a$,and $\int_0^1 x^2 f(x) \, dx = a^2$. Then the value of $\int_0^1 (x - a)^2 f(x) \, dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module