早稲田大学創造理工学研究科経営システム工学専攻 2013年8月実施微积分

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Author

思齐塾, 祭音Myyura

次の定積分 (definite integral) を計算せよ.

I = \int_{0}^{1} \frac{1}{4-x^2} dx

计算定积分

I=\int_{0}^{1}\frac{1}{4-x^2}\,dx.

We can evaluate the integral by using partial fractions. First, we can write

\frac{1}{4-x^2} = \frac{1}{(2-x)(2+x)} = \frac{A}{2-x} + \frac{B}{2+x}

Multiplying both sides by $4-x^2$ , we get

1 = A(2+x) + B(2-x) = (A-B)x + 2A + 2B

Comparing coefficients, we have the following system of equations:

A-B = 0

2A + 2B = 1

From the first equation, $A=B$ . Substituting this into the second equation, we have $4A=1$ , so $A = \frac{1}{4}$ and $B = \frac{1}{4}$ . Thus,

\frac{1}{4-x^2} = \frac{1}{4} \left(\frac{1}{2-x} + \frac{1}{2+x}\right)

Now we can evaluate the integral:

I = \int_0^1 \frac{1}{4-x^2} dx = \frac{1}{4} \int_0^1 \left(\frac{1}{2-x} + \frac{1}{2+x}\right) dx

I = \frac{1}{4} \left[ -\ln|2-x| + \ln|2+x| \right]_0^1 = \frac{1}{4} \left[ \ln\left|\frac{2+x}{2-x}\right| \right]_0^1

I = \frac{1}{4} \left( \ln\left(\frac{2+1}{2-1}\right) - \ln\left(\frac{2+0}{2-0}\right) \right) = \frac{1}{4} \left( \ln(3) - \ln(1) \right) = \frac{1}{4}(\ln 3 - 0) = \frac{1}{4} \ln 3

So, the integral is

I = \frac{\ln 3}{4}

Final Answer: The final answer is $\boxed{\frac{\ln 3}{4}}$