If int_0^3 (3x^2 - 4x + 2) dx = k, then a roo | vedclass

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

Question

(B) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$ $= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$

Vedclass · Accepted Answer

$\frac{7}{3}$

Vedclass · Answer

(B) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$ $= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$. Thus, $k = 15$. Now, we solve the equation $3x^2 - 4x + 2 = \frac{3k}{5}$: $3x^2 - 4x + 2 = \frac{3(15)}{5} = 9$. $3x^2 - 4x - 7 = 0$. Factoring the quadratic equation: $3x^2 - 7x + 3x - 7 = 0$ $x(3x - 7) + 1(3x - 7) = 0$ $(x + 1)(3x - 7) = 0$. The roots are $x = -1$ and $x = \frac{7}{3}$. Since we require the root in the interval $[0, 3]$, the valid root is $x = \frac{7}{3}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

If $\int_0^3 (3x^2 - 4x + 2) dx = k$, then a root of $3x^2 - 4x + 2 = \frac{3k}{5}$ that lies in the interval $[0, 3]$ is

Similar Questions

$\int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x} = $

$\int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx =$

If $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$, then $\int_{0}^{2} f(x) dx$ is

If $y = \sqrt{\sec x + \sqrt{\sec x + \sqrt{\sec x + \dots \infty}}} \,,$ then the value of $\int_{0}^{\pi/3} (2y - 1) \frac{dy}{dx} \, dx$ is equal to $(\sec x > 0)$ -

Using the Trapezoidal rule, find the approximate value of $\int_1^4 y \, dx$ based on the following data:
$x$ $1$ $2$ $3$ $4$
$y$ $0.7111$ $0.7222$ $0.7333$ $0.7444$
(in $.1833$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module