The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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

Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+	an^5 x) dx$. We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+	an^5 x) dx$. We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^5 x dx$. Recall the property of definite integrals: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is an even function. For $f_1(x) = x^3$,$f_1(-x) = (-x)^3 = -x^3 = -f_1(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx = 0$. For $f_2(x) = \cos x$,$f_2(-x) = \cos(-x) = \cos x = f_2(x)$,so it is an even function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx = 2 \int_{0}^{\frac{\pi}{2}} \cos x dx = 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1-0) = 2$. For $f_3(x) = 	an^5 x$,$f_3(-x) = 	an^5(-x) = -	an^5 x = -f_3(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 	an^5 x dx = 0$. Therefore,$I = 0 + 2 + 0 = 2$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$ is equal to . . . . . . .

Similar Questions

The value of $\int_0^\pi \sin^{50} x \cos^{49} x \, dx$ is

For $0 < a < 1$,the value of the integral $\int_0^\pi \frac{d x}{1-2 a \cos x+a^2}$ is:

The value of the definite integral $\int_{19}^{37} (\{x\}^2 + 3 \sin(2\pi x)) \, dx$,where $\{x\}$ denotes the fractional part function.

The option$(s)$ with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4 \pi} e^t(\sin^6 at + \cos^4 at) dt}{\int_0^{\pi} e^t(\sin^6 at + \cos^4 at) dt} = L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module