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 $x = \frac{7}{3}$.

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

If $[\bullet]$ denotes the greatest integer function, then evaluate $\int_0^{2 \pi} [|\sin x| + |\cos x|] \, dx$.

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

Let $f$ be a continuous function in $[0, 1]$, then $\lim_{n \rightarrow \infty} \sum_{j=0}^n \frac{1}{n} f\left(\frac{j}{n}\right)$ is

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

If $f(y) = e^y$,$g(y) = y$ for $y > 0$,and $F(t) = \int_{0}^{t} f(t - y) g(y) dy$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module