$\int_0^{\pi /4} (\cos x - \sin x) dx + \int_{\pi /4}^{5\pi /4} (\sin x - \cos x) dx + \int_{2\pi }^{\pi /4} (\cos x - \sin x) dx$ ની કિંમત શોધો.

  • A
    $\sqrt{2} - 2$
  • B
    $2\sqrt{2} - 2$
  • C
    $3\sqrt{2} - 2$
  • D
    $4\sqrt{2} - 2$

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Similar Questions

જો $\int_{n}^{n+1} g(x) dx = n^2, \forall n \in Z$ હોય,તો $\int_{-3}^3 g(x) dx$ ની કિંમત શોધો.

જો ${I_1} = \int_0^1 {2^{x^2}} dx$,${I_2} = \int_0^1 {2^{x^3}} dx$,${I_3} = \int_1^2 {2^{x^2}} dx$,અને ${I_4} = \int_1^2 {2^{x^3}} dx$ હોય,તો નીચેનામાંથી કયું સાચું છે?

ધારો કે $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. ધ્યાનમાં લો:
$(S1): f^{\prime}\left(-\frac{3}{2}\right) + f^{\prime}\left(-\frac{1}{2}\right) + f^{\prime}\left(\frac{1}{2}\right) + f^{\prime}\left(\frac{3}{2}\right) = 4$
$(S2): \int_{-2}^{2} f(x) dx = 12$
તો,

$ \int_{0}^{2} [x^{2}] \, dx $

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

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