নির্দিষ্ট যোগজ
∫01e−(x2)dx=? \int_{0}^{1} e^{- \left ( x^{2} \right )} dx = ? ∫01e−(x2)dx=?
∫01e−x2dx=∫011−x2+x42!+…=[x−x33+x55.2!…]01=[1−13+15.2!…⋅]=∑k=0∞(−1)k(2k+1)k! \begin{aligned} & \int_{0}^{1} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\int_{0}^{1} 1-\mathrm{x}^{2}+\frac{\mathrm{x}^{4}}{2 !}+\ldots \\ = & {\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{5}}{5.2 !} \ldots\right]_{0}^{1} } \\ = & {\left[1-\frac{1}{3}+\frac{1}{5.2 !} \ldots \cdot\right] } \\ = & \sum_{\mathrm{k}=0}^{\infty} \frac{(-1)^{\mathrm{k}}}{(2 \mathrm{k}+1) \mathrm{k} !}\end{aligned} ===∫01e−x2dx=∫011−x2+2!x4+…[x−3x3+5.2!x5…]01[1−31+5.2!1…⋅]k=0∑∞(2k+1)k!(−1)k
The value of ∫−π/2199π/2(1+cos2x)dx\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx∫−π/2199π/2(1+cos2x)dx is?
∫0π6sin2xcosxdx= \int_{0}^{\frac{\pi}{6}} \sin ^{2} x \cos x d x= ∫06πsin2xcosxdx= কত ?
∫1/21dxx4x2−1\int_{1 / 2}^{1} \frac{d x}{x \sqrt{4 x^{2}-1}}∫1/21x4x2−1dx
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?
