The value of the integral int_{-2}^0 (x^3 + 3 | vedclass

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

Question

(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$. We can rewrite the expression as: $I = \int_{-2}^0 ((x + 1)^3 + 4 + (x +

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$. We can rewrite the expression as: $I = \int_{-2}^0 ((x + 1)^3 + 4 + (x + 1) \cos(x + 1)) \, dx$. Substitute $t = x + 1$,then $dt = dx$. When $x = -2$,$t = -1$. When $x = 0$,$t = 1$. So,$I = \int_{-1}^1 (t^3 + 4 + t \cos t) \, dt$. We know that $t^3$ and $t \cos t$ are odd functions,and the integral of an odd function over a symmetric interval $[-a, a]$ is $0$. Therefore,$I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 4 \, dt + \int_{-1}^1 t \cos t \, dt$. $I = 0 + \int_{-1}^1 4 \, dt + 0$. $I = 4 [t]_{-1}^1 = 4(1 - (-1)) = 4(2) = 8$. The correct value is $8$.

The value of the integral $\int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$ is equal to:

Similar Questions

$\int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x=$

Let $f(x)$ and $g(x)$ be two functions satisfying $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) + g(x) = 0$. Then the value of $\int_{-4}^{4} f(x) dx$ is

The value of $\frac{8}{\pi} \int \limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} dx$ is $.............$.

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

The value of the integral $\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module