If int_a^b x^3 dx = 0 and int_a^b x^2 dx = fr | vedclass

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

Question

(D) Given: $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.Step $1$: Evaluate the first integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \imp

Vedclass · Accepted Answer

$-1, 1$

Vedclass · Answer

(D) Given: $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.Step $1$: Evaluate the first integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \implies \frac{b^4 - a^4}{4} = 0 \implies b^4 = a^4 \implies b^2 = a^2$ or $b = -a$.Step $2$: Evaluate the second integral: $\left[ \frac{x^3}{3} \right]_a^b = \frac{2}{3} \implies \frac{b^3 - a^3}{3} = \frac{2}{3} \implies b^3 - a^3 = 2$.Step $3$: Substitute $b = -a$ into the second equation: $(-a)^3 - a^3 = 2 \implies -a^3 - a^3 = 2 \implies -2a^3 = 2 \implies a^3 = -1 \implies a = -1$.Step $4$: Since $b = -a$,we get $b = -(-1) = 1$.Thus,$a = -1$ and $b = 1$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

If $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$,then $a$ and $b$ are respectively

Similar Questions

$I = \int_{\pi/4}^{\pi/3} \frac{8 \sin x - \sin 2x}{x} dx$. Then

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$,using the Trapezoidal rule,is:
$t \ (\text{in second})$ $0$ $2$ $4$ $6$ $8$ $10$
$v \ (\text{in m/s})$ $0$ $12$ $16$ $20$ $35$ $60$

$a > 1, \; \int_{1}^{a} [x] f'(x) dx = $

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

$\int_2^3 \frac{d x}{x^2-x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module