## 数学の問題

## 数学の問題

与えられた8つの定積分の値を求める問題です。

1. $\int_{0}^{2} x^2 e^{2x} dx$

2. $\int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} \quad (a>0)$

3. $\int_{0}^{\pi/4} \cos^3 x \, dx$

4. $\int_{0}^{\sqrt{3}} \frac{1+x}{1+x^2} \, dx$

5. $\int_{0}^{1} \frac{x}{x^4+1} \, dx$

6. $\int_{0}^{2} \frac{x}{(1+x^2)^2} \, dx$

7. $\int_{0}^{2} x \log(x^2+1) \, dx$

8. $\int_{1}^{2} x e^{x^2} \, dx$

## 解き方の手順

一つずつ積分を計算します。

### (1) $\int_{0}^{2} x^2 e^{2x} dx$

部分積分を2回適用します。

$I = \int x^2 e^{2x} dx$

$u = x^2$, $dv = e^{2x} dx$とすると、$du = 2x dx$, $v = \frac{1}{2} e^{2x}$となります。

$I = \frac{1}{2}x^2 e^{2x} - \int x e^{2x} dx$

次に、$\int x e^{2x} dx$を部分積分します。

$u = x$, $dv = e^{2x} dx$とすると、$du = dx$, $v = \frac{1}{2} e^{2x}$となります。

$\int x e^{2x} dx = \frac{1}{2}x e^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{1}{2}x e^{2x} - \frac{1}{4} e^{2x}$

よって、

$I = \frac{1}{2}x^2 e^{2x} - (\frac{1}{2}x e^{2x} - \frac{1}{4} e^{2x}) = \frac{1}{2}x^2 e^{2x} - \frac{1}{2}x e^{2x} + \frac{1}{4} e^{2x}$

$\int_{0}^{2} x^2 e^{2x} dx = \left[\frac{1}{2}x^2 e^{2x} - \frac{1}{2}x e^{2x} + \frac{1}{4} e^{2x}\right]_0^2$

$= (\frac{1}{2}(4)e^4 - \frac{1}{2}(2)e^4 + \frac{1}{4}e^4) - (0 - 0 + \frac{1}{4})$

$= (2e^4 - e^4 + \frac{1}{4}e^4) - \frac{1}{4} = e^4 + \frac{1}{4}e^4 - \frac{1}{4} = \frac{5e^4 - 1}{4}$

### (2) $\int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} \quad (a>0)$

$x = a \sin\theta$と置換すると、$dx = a \cos\theta \, d\theta$。

$x=0$のとき、$\theta=0$。$x=a/2$のとき、$\sin\theta = 1/2$なので、$\theta = \pi/6$。

$\int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} = \int_0^{\pi/6} \frac{a \cos\theta}{\sqrt{a^2 - a^2 \sin^2\theta}} \, d\theta = \int_0^{\pi/6} \frac{a \cos\theta}{a \cos\theta} \, d\theta = \int_0^{\pi/6} d\theta = [\theta]_0^{\pi/6} = \frac{\pi}{6}$

### (3) $\int_{0}^{\pi/4} \cos^3 x \, dx$

$\cos^3 x = \cos^2 x \cdot \cos x = (1-\sin^2 x) \cos x$

$u = \sin x$と置くと、$du = \cos x \, dx$

$x=0$のとき、$u=0$。$x=\pi/4$のとき、$u=\sin(\pi/4)=\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$。

$\int_{0}^{\pi/4} \cos^3 x \, dx = \int_0^{\sqrt{2}/2} (1-u^2) \, du = [u - \frac{u^3}{3}]_0^{\sqrt{2}/2} = (\frac{\sqrt{2}}{2} - \frac{(\sqrt{2}/2)^3}{3}) - 0 = \frac{\sqrt{2}}{2} - \frac{2\sqrt{2}}{8 \cdot 3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{12} = \frac{6\sqrt{2} - \sqrt{2}}{12} = \frac{5\sqrt{2}}{12}$

### (4) $\int_{0}^{\sqrt{3}} \frac{1+x}{1+x^2} \, dx$

$\int_{0}^{\sqrt{3}} \frac{1+x}{1+x^2} \, dx = \int_{0}^{\sqrt{3}} \frac{1}{1+x^2} \, dx + \int_{0}^{\sqrt{3}} \frac{x}{1+x^2} \, dx$

$\int_{0}^{\sqrt{3}} \frac{1}{1+x^2} \, dx = [\arctan x]_0^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3}$

$\int_{0}^{\sqrt{3}} \frac{x}{1+x^2} \, dx$で、$u = 1+x^2$と置換すると、$du = 2x \, dx$

$x=0$のとき、$u=1$。$x=\sqrt{3}$のとき、$u=1+3=4$。

$\int_{0}^{\sqrt{3}} \frac{x}{1+x^2} \, dx = \frac{1}{2}\int_1^4 \frac{1}{u} \, du = \frac{1}{2}[\ln u]_1^4 = \frac{1}{2}(\ln 4 - \ln 1) = \frac{1}{2}\ln 4 = \ln 2$

$\int_{0}^{\sqrt{3}} \frac{1+x}{1+x^2} \, dx = \frac{\pi}{3} + \ln 2$

### (5) $\int_{0}^{1} \frac{x}{x^4+1} \, dx$

$u = x^2$と置換すると、$du = 2x \, dx$

$x=0$のとき、$u=0$。$x=1$のとき、$u=1$。

$\int_{0}^{1} \frac{x}{x^4+1} \, dx = \frac{1}{2} \int_0^1 \frac{1}{u^2+1} \, du = \frac{1}{2} [\arctan u]_0^1 = \frac{1}{2} (\arctan(1) - \arctan(0)) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$

### (6) $\int_{0}^{2} \frac{x}{(1+x^2)^2} \, dx$

$u = 1+x^2$と置換すると、$du = 2x \, dx$

$x=0$のとき、$u=1$。$x=2$のとき、$u=1+4=5$。

$\int_{0}^{2} \frac{x}{(1+x^2)^2} \, dx = \frac{1}{2} \int_1^5 \frac{1}{u^2} \, du = \frac{1}{2} \int_1^5 u^{-2} \, du = \frac{1}{2} [-u^{-1}]_1^5 = \frac{1}{2} [-\frac{1}{u}]_1^5 = \frac{1}{2} (-\frac{1}{5} - (-1)) = \frac{1}{2} (1 - \frac{1}{5}) = \frac{1}{2} (\frac{4}{5}) = \frac{2}{5}$

### (7) $\int_{0}^{2} x \log(x^2+1) \, dx$

$u = x^2+1$と置換すると、$du = 2x \, dx$

$x=0$のとき、$u=1$。$x=2$のとき、$u=4+1=5$。

$\int_{0}^{2} x \log(x^2+1) \, dx = \frac{1}{2} \int_1^5 \log u \, du$

$\int \log u \, du = u \log u - u$ (部分積分)

$\frac{1}{2} \int_1^5 \log u \, du = \frac{1}{2} [u \log u - u]_1^5 = \frac{1}{2} [(5 \log 5 - 5) - (1 \log 1 - 1)] = \frac{1}{2} (5 \log 5 - 5 - 0 + 1) = \frac{1}{2} (5 \log 5 - 4)$

### (8) $\int_{1}^{2} x e^{x^2} \, dx$

$u = x^2$と置換すると、$du = 2x \, dx$

$x=1$のとき、$u=1$。$x=2$のとき、$u=4$。

$\int_{1}^{2} x e^{x^2} \, dx = \frac{1}{2} \int_1^4 e^u \, du = \frac{1}{2} [e^u]_1^4 = \frac{1}{2} (e^4 - e^1) = \frac{e^4 - e}{2}$

## 最終的な答え

1. $\frac{5e^4 - 1}{4}$

2. $\frac{\pi}{6}$

3. $\frac{5\sqrt{2}}{12}$

4. $\frac{\pi}{3} + \ln 2$

5. $\frac{\pi}{8}$

6. $\frac{2}{5}$

7. $\frac{5 \log 5 - 4}{2}$

8. $\frac{e^4 - e}{2}$