নির্দিষ্ট যোগজ
∫012dx1−3x2=? \int_{0}^{\frac{1}{2}} \frac{dx}{\sqrt{1 - 3 x^{2}}} = ? ∫0211−3x2dx=?
π/2
2π/3
∫01/211−3x2dx=13∫0121(13)2−x2dx=[13sin−1(3x)]012=13⋅π3∴I=π33 \begin{array}{l}\text { } \int_{0}^{1 / 2} \frac{1}{\sqrt{1-3 \mathrm{x}^{2}}} \mathrm{dx} \\ =\frac{1}{\sqrt{3}} \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}-\mathrm{x}^{2}}} \mathrm{dx}=\left[\frac{1}{\sqrt{3}} \sin ^{-1}(\sqrt{3} \mathrm{x})\right]_{0}^{\frac{1}{2}}=\frac{1}{\sqrt{3}} \cdot \frac{\pi}{3} \quad \therefore \mathrm{I}=\frac{\pi}{3 \sqrt{3}}\end{array} ∫01/21−3x21dx=31∫021(31)2−x21dx=[31sin−1(3x)]021=31⋅3π∴I=33π
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
