ক্ষেত্রফল নির্ণয়

ক) x(x1)(x2+1)dx \int \frac{x}{\left ( x - 1 \right ) \left ( x^{2} + 1 \right )} dx

 খ) y² =16x এবং y = x দ্বারা সীমাবদ্ধ ক্ষেত্রের ক্ষেত্রফল কত?

BUTEX 00-01

ক)

 12dx(x1)+12dx(x2+1)12x(x2+1)dx ধরি, x(x1)(x2+1)=12ln(x1)+12tan1x142x(x2+1)dx=Ax1+Bx+Cx2+1=12ln(x1)+12tan1x14ln(x2+1)+c (Ans) x=A(x2+1)+(B B2CBx+Cx)0.x2+x=(A+B)x2(BC)x+AC \begin{array}{l|l} \text { } & \therefore \frac{1}{2} \int \frac{\mathrm{dx}}{(\mathrm{x}-1)}+\frac{1}{2} \int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+1\right)}-\frac{1}{2} \int \frac{\mathrm{x}}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx} \\ \text { ধরি, } \frac{\mathrm{x}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)} & =\frac{1}{2} \ln (\mathrm{x}-1)+\frac{1}{2} \tan ^{-1} \mathrm{x}-\frac{1}{4} \int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx} \\ =\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+1} & =\frac{1}{2} \ln (\mathrm{x}-1)+\frac{1}{2} \tan ^{-1} \mathrm{x}-\frac{1}{4} \ln \left(\mathrm{x}^{2}+1\right)+\mathrm{c} \text { (Ans) } \\ \therefore \mathrm{x}=\mathrm{A}\left(\mathrm{x}^{2}+1\right)+\left(B \mathrm{~B}^{2}-\mathrm{C}-\mathrm{Bx}+\mathrm{Cx}\right) & 0 . \mathrm{x}^{2}+\mathrm{x}=(\mathrm{A}+\mathrm{B}) \mathrm{x}^{2}-(\mathrm{B}-\mathrm{C}) \mathrm{x}+\mathrm{A}-\mathrm{C} \end{array}

x2 \mathrm{x}^{2} সহগ সমতাকৃত করে পাই, 0=A+BA=B 0=\mathrm{A}+\mathrm{B} \quad \therefore \mathrm{A}=-\mathrm{B}

x \mathrm{x} সহগ সমতাকৃত করে পাই, (BC)=1B=C1 -(\mathrm{B}-\mathrm{C})=1 \Rightarrow \mathrm{B}=\mathrm{C}-1 x=1 \mathrm{x}=1 বসিয়ে পাই A=12 \mathrm{A}=\frac{1}{2}

B=12C=12 \therefore B=-\frac{1}{2} \therefore C=\frac{1}{2}

খ) ক্ষেত্রফল =016((4x)x)dx \left.=\int_{0}^{16}((4 \sqrt{\mathrm{x}})-\mathrm{x}\right) \mathrm{dx}

y2=16xx(x16)=0x=0,16 \begin{array}{l} y^{2}=16 x \\ \Rightarrow x(x-16)=0 \\ \therefore x=0,16 \end{array}

=4(23)[x3/2]01612[x2]016=83×642562=1283 =4\left(\frac{2}{3}\right)\left[\mathrm{x}^{3 / 2}\right]_{0}^{16}-\frac{1}{2}\left[\mathrm{x}^{2}\right]_{0}^{16}=\frac{8}{3} \times 64-\frac{256}{2}=\frac{128}{3} বর্গ একক (Ans.)

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