নির্দিষ্ট যোগজ
∫23dx(x−1)x2−2x \int_{2}^{3} \frac{dx}{\left ( x - 1 \right ) \sqrt{x^{2} - 2 x}} ∫23(x−1)x2−2xdx এর মান__
∫23dx(x−1)x2−2x \int_{2}^{3} \frac{\mathrm{dx}}{(\mathrm{x}-1) \sqrt{\mathrm{x}^{2}-2 \mathrm{x}}} \quad ∫23(x−1)x2−2xdx । Let, x2−2x=z2∴2zdz=(2x−2)dx∴zdz=(x−1)dx \mathrm{x}^{2}-2 \mathrm{x}=\mathrm{z}^{2} \therefore 2 \mathrm{zdz}=(2 \mathrm{x}-2) \mathrm{dx} \therefore \mathrm{zdz}=(\mathrm{x}-1) \mathrm{dx} x2−2x=z2∴2zdz=(2x−2)dx∴zdz=(x−1)dx
=∫23(x−1)(x2−2x+1)x2−2x=∫23zdz(z2+1)z2=∫23dzz2+1=[11tan−1z]23=[tan−1x2−2x]23=tan−13−tan−10=tan−1tanπ3=π3 \begin{array}{l} =\int_{2}^{3} \frac{(x-1)}{\left(x^{2}-2 x+1\right) \sqrt{x^{2}-2 x}}=\int_{2}^{3} \frac{z d z}{\left(z^{2}+1\right) \sqrt{z^{2}}}=\int_{2}^{3} \frac{d z}{z^{2}+1}=\left[\frac{1}{1} \tan ^{-1} z\right]_{2}^{3}=\left[\tan ^{-1} \sqrt{x^{2}-2 x}\right]_{2}^{3} \\ =\tan ^{-1} \sqrt{3}-\tan ^{-1} 0=\tan ^{-1} \tan \frac{\pi}{3}=\frac{\pi}{3} \end{array} =∫23(x2−2x+1)x2−2x(x−1)=∫23(z2+1)z2zdz=∫23z2+1dz=[11tan−1z]23=[tan−1x2−2x]23=tan−13−tan−10=tan−1tan3π=3π
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
