ক্ষেত্রফল নির্ণয়

Let f(x,y)={(x,y):y24x,0xλ}f(x, y)=\{(x, y): y^2 \leq 4x, 0\leq x\leq \lambda\} and s(λ)s(\lambda) is area such that S(λ)S(4)=25\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}. Find the value of λ\lambda.

হানি নাটস

y2=4xy^2=4x

S(I)=20λ2xdx=4x3/23/2]0λ=83λ3/2S(I)=\left.2\displaystyle\int^{\lambda}_{0}2\sqrt{x}dx=\dfrac{4x^{3/2}}{3/2}\right]^{\lambda}_{0}=\dfrac{8}{3}\lambda^{3/2}

S(λ)S(4)=25\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5} λ3/243/2=25\Rightarrow \dfrac{\lambda^{3/2}}{4^{3/2}}=\dfrac{2}{5}

λ=4(25)2/3=4(425)1/3\lambda =4\left(\dfrac{2}{5}\right)^{2/3}=4\left(\dfrac{4}{25}\right)^{1/3}.

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