নির্দিষ্ট যোগজ
∫π3π2cos5xsin7xdx \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos^{5}{x}}{\sin^{7}{x}} dx ∫3π2πsin7xcos5xdx এর মান কোনটি ?
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−∫π3π2cot5xd(cotx)=−16[cot6x]π3π2=−16[0−1(3)6]=1162 \begin{array}{l}\text { }-\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cot ^{5} \mathrm{x} d(\cot x) \\ =-\frac{1}{6}\left[\cot ^{6} \mathrm{x}\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=-\frac{1}{6}\left[0-\frac{1}{(\sqrt{3})^{6}}\right]=\frac{1}{162}\end{array} −∫3π2πcot5xd(cotx)=−61[cot6x]3π2π=−61[0−(3)61]=1621
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
দৃশ্যকল্প-১: f(x)=x2tan−1x31+x6 f(\mathrm{x})=\frac{\mathrm{x}^{2} \tan ^{-1} \mathrm{x}^{3}}{1+\mathrm{x}^{6}} f(x)=1+x6x2tan−1x3
দৃশ্যকল্প-২: g(x,y)=16x2+25y2−400 g(x, y)=16 x^{2}+25 y^{2}-400 g(x,y)=16x2+25y2−400
