নির্দিষ্ট যোগজ
∫1e2dxx(1+lnx) \int_{1}^{e^{2}} \frac{dx}{x \left ( 1 + \ln{x} \right )} ∫1e2x(1+lnx)dx এর মান কত?
ln3
ln2
1
0
∫1e2dxx(1+lnx) Let, 1+lnx=z∴1xdx=dz=∫1e21zdz=[ln(1+lnx)]1ev=ln3. \begin{array}{l}\begin{array}{l|l}\int_{1}^{e^{2}} \frac{d x}{x(1+\ln x)} & \begin{array}{l}\text { Let, } \\ 1+\ln x=z \\ \therefore \frac{1}{x} d x=d z\end{array}\end{array} \\ =\int_{1}^{e^{2}} \frac{1}{z} d z \\ =[\ln (1+\ln x)]_{1}^{e^{v}} \\ =\ln 3 . \\\end{array} ∫1e2x(1+lnx)dx Let, 1+lnx=z∴x1dx=dz=∫1e2z1dz=[ln(1+lnx)]1ev=ln3.
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?
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
f(x)=sinxf(x)=\sin xf(x)=sinx
g(x,y)=9x2+25y2−225g(x, y)=9 x^{2}+25 y^{2}-225g(x,y)=9x2+25y2−225
h(x)=x−3h(x)=x-3h(x)=x−3
