1.次の不定積分を求めなさい。
(1)\(\displaystyle \int(x^3-6x^2-2x+5)dx\)
\(\displaystyle =\frac{1}{4}x^4-2x^3-x^2+5x+C\ \ \)(\(C\)は積分定数)
(2)\(\displaystyle \int(x+1)(x-3)dx\)
\(\displaystyle =\int(x^2-2x-3)dx\)
\(\displaystyle =\frac{1}{3}x^3-x^2-3x+C\ \ \)(\(C\)は積分定数)
(3)\(\displaystyle \int3(t-1)^2dx\)
\(\displaystyle =\int(3t^2-6t+3)dx\)
\(\displaystyle =t^3-3t^2+3t+C\ \ \)(\(C\)は積分定数)
(4)\(\displaystyle \int(3x-2)dx\)
\(\displaystyle =\frac{3}{2}x^2-2x+C\ \ \)(\(C\)は積分定数)
(5)\(\displaystyle \int(-3x^2+4x+1)dx\)
\(=-x^3+2x^2+x+C\ \ \)(\(C\)は積分定数)
(6)\(\displaystyle \int(2t-1)(2t+1)dx\)
\(\displaystyle =\int(4t^2-1)dx\)
\(\displaystyle =\frac{4}{3}t^3-t+C\ \ \)(\(C\)は積分定数)
(7)\(\displaystyle \int(2x+1)dx\)
\(=x^2+x+C\ \ \)(\(C\)は積分定数)
(8)\(\displaystyle \int(x^2-3x-5)dx\)
\(\displaystyle =\frac{1}{3}x^3-\frac{3}{2}x^2-5x+C\ \ \)(\(C\)は積分定数)
(9)\(\displaystyle \int(2t^2+1)(2t-3)dx\)
\(\displaystyle =\int(4t^3-6t^2+2t-3)dx\)
\(=t^4-2t^3+t^2-3t+C\ \ \)(\(C\)は積分定数)
2.次の条件をみたす関数\(f(x)\)を求めなさい。
(1)\(f'(x)=3x^2-4,f(1)=2\)
\(\displaystyle f(x)=\int(3x^2-4)dx\)
\(\ \ \ \ \ \ \ =x^3-4x+C\)
\(f(1)=2\)より、
\(2=1-4+C\)
\(C=5\)
よって、
\(f(x)=x^3-4x+5\)
(2)\(f'(x)=x^2+5,f(3)=30\)
\(\displaystyle f(x)=\int(x^2+5)dx\)
\(\displaystyle \ \ \ \ \ \ \ =\frac{1}{3}x^3+5x+C\)
\(f(3)=30\)より、
\(30=9+15+C\)
\(C=6\)
よって、
\(\displaystyle f(x)=\frac{1}{3}x^3+5x+6\)
(3)\(f'(x)=3x^2-x,f(2)=7\)
\(\displaystyle f(x)=\int(3x^2-x)dx\)
\(\displaystyle \ \ \ \ \ \ \ =x^3-\frac{1}{2}x^2+C\)
\(f(2)=7\)より、
\(7=8-2+C\)
\(C=1\)
よって、
\(\displaystyle f(x)=x^3-\frac{1}{2}x^2+1\)