নির্দিষ্ট যোগজ
f(x)=ex, g(x)=sinx+cosx
∫0π4g(x)dx \int_{0}^{\frac{\pi}{4}} g{\left ( x \right )} dx ∫04πg(x)dx এর মান নিচের কোনটি ?
1
∫0π4(sinx+cosx)dx \int_{0}^{\frac{\pi}{4}}(\sin x+\cos x) d x ∫04π(sinx+cosx)dx
=[−cosx+sinx)]0π4=(−12+12)−(−1+0)=1 \begin{array}{l} =[-\cos x+\sin x)]_{0}^{\frac{\pi}{4}} \\ =\left(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(-1+0)=1 \end{array} =[−cosx+sinx)]04π=(−21+21)−(−1+0)=1
∫0π/2cosxdx= কত? \int_{0}^{\pi / 2} \cos x d x=\text { কত? } ∫0π/2cosxdx= কত?
f(x)= {x+1forx=0 \left \lbrace \begin{matrix} x + 1 & f{\quad\text{or}\quad} & x & = & 0 \end{matrix} \right . {x+1forx=0 হলে-
∫−1−12f(x)dx=18 \int_{- 1}^{- \frac{1}{2}} f{\left ( x \right )} dx = \frac{1}{8} ∫−1−21f(x)dx=81
∫01f(x)dx=0 \int_{0}^{1} f{\left ( x \right )} dx = 0 ∫01f(x)dx=0
f(−1)=1 f{\left ( - 1 \right )} = 1 f(−1)=1
নিচের কোনটি সঠিক?
∫1e2dxx(1+lnx) \int_{1}^{e^{2}} \frac{dx}{x \left ( 1 + \ln{x} \right )} ∫1e2x(1+lnx)dx এর মান কত?
α এর মান কত হলে ∫1α{2+xln(x2+5)}dx+∫1α{3−xln(x2+5)}dx \int_{1}^{\alpha} \left \lbrace 2 + x \ln{\left ( x^{2} + 5 \right )} \right \rbrace dx + \int_{1}^{\alpha} \left \lbrace 3 - x \ln{\left ( x^{2} + 5 \right )} \right \rbrace dx ∫1α{2+xln(x2+5)}dx+∫1α{3−xln(x2+5)}dx =30