int_{-2}^0 (x^3+3x^2+3x+3+(x+1) cos(x+1)) , d | vedclass

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

Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $I = \int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$. We can rewrite the integrand as: $I = \int_{-2}^0 ((x+1)^3 + 2 + (x+1) \cos(x+1)) \, dx$. Substitute $t = x+1$,so $dt = dx$. When $x = -2$,$t = -1$. When $x = 0$,$t = 1$. Thus,$I = \int_{-1}^1 (t^3 + 2 + t \cos t) \, dt$. We can split the integral: $I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 2 \, dt + \int_{-1}^1 t \cos t \, dt$. Since $f(t) = t^3$ and $g(t) = t \cos t$ are odd functions,their integrals over the symmetric interval $[-1, 1]$ are $0$. Therefore,$I = 0 + [2t]_{-1}^1 + 0 = 2(1 - (-1)) = 2(2) = 4$.

$\int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$ is equal to:

Similar Questions

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

If $f(x) = \frac{e^x}{1 + e^x}$,$I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$,and $I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$,then the value of $\frac{I_2}{I_1}$ is

If $f$ is defined on $R$ such that $f(x) f(-x) = 9$, then find the value of $\int_{-23}^{23} \frac{dx}{3+f(x)}$.

The value of $\int_{-2}^{2} \left[ p \ln \left( \frac{1+x}{1-x} \right) + q \ln \left( \frac{1-x}{1+x} \right)^{-2} + r \right] dx$ depends on:

If $I_{n}=\int_0^{\frac{\pi}{4}} \tan ^n x \, dx$,then $I_{13}+I_{11}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module