積分の公式を使った問題

1. 次の不定積分を解きなさい．

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  ${\displaystyle \int \left(x^3+2x^2-\sqrt{x}\right)dx}$

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  ${\displaystyle \int \frac{3}{\sqrt{x+1}-\sqrt{x}}dx}$

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  ${\displaystyle \int \frac{3x^3+12x+1}{x^2+4}dx}$

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

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  ${\displaystyle \int \frac{1}{\sqrt{4-9x^2}}dx}$

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  ${\displaystyle \int \frac{1}{{\left(3x+1\right)}^2+4}dx}$

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  ${\displaystyle \int \frac{1}{\sqrt{x^2+5}}dx}$

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  ${\displaystyle \int 5^{x}dx}$

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  ${\displaystyle \int \frac{1}{{\cos }^{2}2x}dx}$

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  ${\displaystyle {\displaystyle \int {\tan }^{2}2x}dx}$

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  ${\displaystyle \int sin3xcos5xdx}$

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  ${\displaystyle {\displaystyle \int \sin 3x\sin 5x}dx}$

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  ${\displaystyle {\displaystyle \int \cos 3x\cos 5x}dx}$

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  ${\displaystyle {\displaystyle \int {\sin }^{2}2x}dx}$

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  ${\displaystyle {\displaystyle \int {\cos }^{2}2x}dx}$

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  ${\displaystyle {\displaystyle \int \frac{1}{{\tan }^{2}2x}}dx}$

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  ${\displaystyle \int {\left(3x^2-5x+2\right)}^2\left(6x-5\right)dx}$

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  ${\displaystyle \int x\sqrt{3x-5}dx}$

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  ${\displaystyle \int \frac{1}{3x-1}dx}$

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  ${\displaystyle {\displaystyle \int \frac{6x-5}{3x^2-5x+2}}dx}$

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  ${\displaystyle \int e^{3x}dx}$

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

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  ${\displaystyle {\displaystyle \int \frac{\log 2x}{x}}dx}$

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  ${\displaystyle \int sin\left(5x+\frac{\pi }{3}\right)dx}$

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  ${\displaystyle \int cos\left(5x+\frac{\pi }{3}\right)xdx}$

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  ${\displaystyle \int tan2xdx}$

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  ${\displaystyle \int {sec}^2\left(5x+\frac{\pi }{3}\right)dx}$

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  ${\displaystyle \int {\text{cosec}}^2\left(5x+\frac{\mathrm{\pi }}{3}\right)dx}$

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  ${\displaystyle \int \frac{x}{2x+1}dx}$

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  ${\displaystyle {\displaystyle \int \sin 2x\sin xdx}}$

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  ${\displaystyle {\displaystyle \int x\sqrt{x+1}dx}}$

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  ${\displaystyle {\displaystyle \int \frac{1}{\sqrt{x+2}-\sqrt{x-2}}dx}}$

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  ${\displaystyle {\displaystyle \int {\tan }^{3}x{\sec }^{2}x}dx}$

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  ${\displaystyle {\displaystyle \int \frac{{\sec }^{2}x}{{\tan }^{3}x}}dx}$

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  ${\displaystyle {\displaystyle \int e^{\cos x}\sin x}dx}$

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  ${\displaystyle {\displaystyle \int \left(5x^4-3x^2+\frac{3\sqrt{x}}{2}\right)}dx}$

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  ${\displaystyle {\displaystyle \int \frac{1}{\sqrt{1-4x^2}}}dx}$

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2. 次の定積分を解きなさい．

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  ${\displaystyle {\int }_1^2\left(x^3-3x^2+\frac{1}{\sqrt{x}}\right)dx}$

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  ${\displaystyle {\int }_{\frac{1}{2}}^4\frac{1}{8x+3}dx}$

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  ${\displaystyle {\int }_2^4{e}^{\frac{1}{2}x}dx}$

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  ${\displaystyle {\int }_{\frac{1}{3}}^2{8}^{x}dx}$

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  ${\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}sin\left(x+\frac{\pi }{2}\right)dx}$

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  ${\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}cos\left(x+\frac{\pi }{2}\right)dx}$

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  ${\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}{sec}^2\left(x+\frac{\pi }{2}\right)dx}$

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  ${\displaystyle {\int }_{-\frac{\pi }{4}}^{-\frac{\pi }{6}}{\text{cosec}}^2\left(x+\frac{\pi }{2}\right)dx}$

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  ${\displaystyle {\int }_1^{\sqrt{3}}\frac{1}{{\left(x+1\right)}^2+3}dx}$

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  ${\displaystyle {\int }_1^{\sqrt{2}}\frac{1}{\sqrt{4-2x^2}}dx}$

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  ${\displaystyle {\int }_{-3}^0\frac{1}{\sqrt{x^2+16}}dx}$

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  ${\displaystyle {\displaystyle {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{\sin x}}dx}$

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  ${\displaystyle {{\displaystyle \int }}_0^1{\left(x\sqrt{1-x^2}\right)}^{3}dx}$

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  ${\displaystyle {\displaystyle {\int }_0^1\frac{x^2+1}{x+1}dx}}$

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  ${\displaystyle {\displaystyle {\int }_0^{\frac{\pi }{4}}\sin x\cos xdx}}$

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  ${\displaystyle {\displaystyle {\int }_1^{e}x\log xdx}}$

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  ${\displaystyle {\displaystyle {\int }_1^3\frac{4x}{1+x^2}}dx}$

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  ${\displaystyle {\displaystyle {\int }_{\frac{1}{4}}^1{16}^{x}dx}}$

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  ${\displaystyle {\displaystyle {\int }_0^1{x}^2{e}^x}dx}$

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  ${\displaystyle {\displaystyle {\int }_0^1\frac{x}{\sqrt{1-x^2}}}dx}$

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  ${\displaystyle {\displaystyle {\int }_0^1\frac{x}{{\left(1-x^2\right)}^{\frac{1}{3}}}}dx}$

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  ${\displaystyle {\displaystyle {\int }_{-3}^{-4}x{(x+3)}^2}dx}$

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  ${\displaystyle {\displaystyle {\int }_0^1\sqrt{4-x^2}}dx}$

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  ${\displaystyle {\displaystyle {\int }_0^1{x}^2{e}^{x^3}}dx}$

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3. 次の不定積分を部分積分法を用いて解きなさい．

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  ${\displaystyle \int x^3{e}^{x}dx}$

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  ${\displaystyle \int 2x{e}^{2x}dx}$

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  ${\displaystyle \int xsinxdx}$

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  ${\displaystyle \int xcosxdx}$

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  ${\displaystyle \int e^{2x}sinxdx}$

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  ${\displaystyle \int \sqrt{x^2+5}dx}$

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  ${\displaystyle \int log2xdx}$

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  ${\displaystyle \int log\left(x+1\right)dx}$

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  ${\displaystyle \int 2log\left(2x+1\right)dx}$

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  ${\displaystyle \int {\left(log2x\right)}^{3}dx}$

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  ${\displaystyle {\displaystyle \int {\tan }^{-1}x}dx}$

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  ${\displaystyle {\displaystyle \int x\log 3xdx}}$

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  ${\displaystyle {\displaystyle \int {(\log x)}^2}dx}$

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4. 次の不定積分を解きなさい．

•	${\displaystyle \int \frac{1}{cosx}dx}$ 　を ${\displaystyle tan\frac{x}{2}=t}$ 　で置換して解きなさい．	⇒解答
•	${\displaystyle \int \frac{1}{sinx}dx}$ 　を ${\displaystyle tan\frac{x}{2}=t}$ 　で置換して解きなさい．	⇒解答
•	${\displaystyle \int \frac{1}{sinx+cosx+1}dx}$ 　を ${\displaystyle tan\frac{x}{2}=t}$ 　で置換して解きなさい．	⇒解答

5. 次の不定積分${\displaystyle \int sinxcosxdx}$ をそれぞれの条件で解きなさい．

•	${\displaystyle 2}$ 倍角の公式を用いて解きなさい．	⇒解答
•	${\displaystyle cosx=t}$ 　で置換して解きなさい．	⇒解答
•	${\displaystyle sinx=t}$ 　で置換して解きなさい．	⇒解答
•	${\displaystyle {\displaystyle \int {\left(-\cos x\right)}^{\prime }\cos x}dx}$ 　と考え部分積分を用いて解きなさい．	⇒解答
•	${\displaystyle \int sinx{\left(sinx\right)}^{\prime }dx}$ 　と考え部分積分を用いて解きなさい．	⇒解答

6. 次の曲線,直線,${\displaystyle x}$軸で囲まれた面積を求めよ.

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  ${\displaystyle y=x^2-2x-3}$ と${\displaystyle x}$軸

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  ${\displaystyle y=x^2-x-1}$ と${\displaystyle y=2x+3}$

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  ${\displaystyle y=3x^2+1}$ ，${\displaystyle x=1}$，${\displaystyle x=3}$

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  ${\displaystyle y=x^2-2x-1}$ と${\displaystyle y=-x^2+x+8}$

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  ${\displaystyle {y=x}^2+4x-3}$ と${\displaystyle {y=-x}^2+2x+1}$

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  ${\displaystyle y=-x^4+2x^3}$ ，${\displaystyle x}$ 軸，${\displaystyle x=1}$

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7.次の図形の重心を求めなさい．

•	直線 ${\displaystyle y=\frac{3}{2}x\left(0\le x\le 2\right)}$ と ${\displaystyle x}$ 軸で囲まれた図形の重心を求めなさい．	⇒解答
•	曲線${\displaystyle y=\frac{3}{4}{x}^2\left(0\le x\le 2\right)}$ と${\displaystyle x}$軸に囲まれた図形の重心を求めなさい．	⇒解答
•	直線${\displaystyle y=\frac{1}{4}x\left(0\le x\le 4\right)}$ と${\displaystyle x}$軸に囲まれた図形の重心を求めなさい．	⇒解答
•	直線${\displaystyle y=x\left(0\le x\le 3\right)}$ と${\displaystyle x}$軸に囲まれた図形の重心を求めなさい．	⇒解答
•	曲線${\displaystyle y=\frac{1}{16}{x}^2\left(0\le x\le 4\right)}$と${\displaystyle x}$軸に囲まれた図形の重心を求めなさい．	⇒解答
•	直線${\displaystyle y=\frac{3}{2}x\left(0\le x\le 2\right)}$ と ${\displaystyle x}$ 軸で囲まれた図形を${\displaystyle x}$軸の周りに${\displaystyle 1}$ 回転してできる立体の重心を求めなさい．	⇒解答
•	${\displaystyle 1\le x\le 4}$ の範囲で，直線${\displaystyle y=x}$ と曲線${\displaystyle y=x^2-4x+4}$ で囲まれた図形の重心を求めなさい．	⇒解答

8. 次の問題を解きなさい．

•  ${\displaystyle y=x+\sqrt{x^2+5}}$ の逆関数を${\displaystyle y=f(x)}$ とする． （１）${\displaystyle f(x)}$ を求めよ． （２）定積分${\displaystyle {\displaystyle {\int }_{\sqrt{5}}^{5}f(x)dx}}$ を求めよ．

⇒解答

•  ${\displaystyle y=x{e}^x(x\ge 0)}$ の逆関数を ${\displaystyle y=f(x)}$ とおく. 定積分 ${\displaystyle {\int }_0^{e}f(x)dx}$ を求めよ．

⇒解答

 

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最終更新日： 2023年1月11日