Question

1. $f(x)=6x + 12x^{2}$
2. $f(x)=2 + x^{3}+x^{6}$
3. $f(x)=\frac{2}{x^{2/3}}$
4. $f(x)=6x+sin x$

Explanation:

1. For \(f''(x)=6x + 12x^{2}\):

• Step 1: Integrate \(f''(x)\) to find \(f'(x)\)
• Using the power - rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n

eq - 1\)).

• \(\int(6x + 12x^{2})dx=\int6xdx+\int12x^{2}dx\).
• \(\int6xdx = 6\times\frac{x^{2}}{2}=3x^{2}\) and \(\int12x^{2}dx=12\times\frac{x^{3}}{3}=4x^{3}\). So, \(f'(x)=3x^{2}+4x^{3}+C_1\).
• Step 2: Integrate \(f'(x)\) to find \(f(x)\)
• \(\int(3x^{2}+4x^{3}+C_1)dx=\int3x^{2}dx+\int4x^{3}dx+\int C_1dx\).
• \(\int3x^{2}dx = 3\times\frac{x^{3}}{3}=x^{3}\), \(\int4x^{3}dx=4\times\frac{x^{4}}{4}=x^{4}\), and \(\int C_1dx=C_1x + C_2\).
• So, \(f(x)=x^{3}+x^{4}+C_1x + C_2\).

1. For \(f''(x)=2 + x^{3}+x^{6}\):

• Step 1: Integrate \(f''(x)\) to find \(f'(x)\)
• \(\int(2 + x^{3}+x^{6})dx=\int2dx+\int x^{3}dx+\int x^{6}dx\).
• \(\int2dx = 2x\), \(\int x^{3}dx=\frac{x^{4}}{4}\), \(\int x^{6}dx=\frac{x^{7}}{7}\). So, \(f'(x)=2x+\frac{x^{4}}{4}+\frac{x^{7}}{7}+C_1\).
• Step 2: Integrate \(f'(x)\) to find \(f(x)\)
• \(\int(2x+\frac{x^{4}}{4}+\frac{x^{7}}{7}+C_1)dx=\int2xdx+\int\frac{x^{4}}{4}dx+\int\frac{x^{7}}{7}dx+\int C_1dx\).
• \(\int2xdx=x^{2}\), \(\int\frac{x^{4}}{4}dx=\frac{1}{4}\times\frac{x^{5}}{5}=\frac{x^{5}}{20}\), \(\int\frac{x^{7}}{7}dx=\frac{1}{7}\times\frac{x^{8}}{8}=\frac{x^{8}}{56}\), \(\int C_1dx=C_1x + C_2\).
• So, \(f(x)=x^{2}+\frac{x^{5}}{20}+\frac{x^{8}}{56}+C_1x + C_2\).

1. For \(f''(x)=\frac{2}{x^{2/3}}\):

• First, rewrite \(f''(x)\) as \(f''(x)=2x^{-\frac{2}{3}}\).
• Step 1: Integrate \(f''(x)\) to find \(f'(x)\)
• Using the power - rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n + 1}+C\) (\(n

eq - 1\)).

• \(\int2x^{-\frac{2}{3}}dx=2\times\frac{x^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}=2\times\frac{x^{\frac{1}{3}}}{\frac{1}{3}} = 6x^{\frac{1}{3}}+C_1\).
• Step 2: Integrate \(f'(x)\) to find \(f(x)\)
• \(\int(6x^{\frac{1}{3}}+C_1)dx=\int6x^{\frac{1}{3}}dx+\int C_1dx\).
• \(\int6x^{\frac{1}{3}}dx=6\times\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}=6\times\frac{x^{\frac{4}{3}}}{\frac{4}{3}}=\frac{9}{2}x^{\frac{4}{3}}\), and \(\int C_1dx=C_1x + C_2\).
• So, \(f(x)=\frac{9}{2}x^{\frac{4}{3}}+C_1x + C_2\).

1. For \(f''(x)=6x+\sin x\):

• Step 1: Integrate \(f''(x)\) to find \(f'(x)\)
• \(\int(6x+\sin x)dx=\int6xdx+\int\sin xdx\).
• \(\int6xdx = 3x^{2}\) and \(\int\sin xdx=-\cos x\). So, \(f'(x)=3x^{2}-\cos x + C_1\).
• Step 2: Integrate \(f'(x)\) to find \(f(x)\)
• \(\int(3x^{2}-\cos x + C_1)dx=\int3x^{2}dx-\int\cos xdx+\int C_1dx\).
• \(\int3x^{2}dx=x^{3}\), \(\int\cos xdx=\sin x\), and \(\int C_1dx=C_1x + C_2\).
• So, \(f(x)=x^{3}-\sin x+C_1x + C_2\).

Answer:

• For \(f''(x)=6x + 12x^{2}\), \(f(x)=x^{3}+x^{4}+C_1x + C_2\).
• For \(f''(x)=2 + x^{3}+x^{6}\), \(f(x)=x^{2}+\frac{x^{5}}{20}+\frac{x^{8}}{56}+C_1x + C_2\).
• For \(f''(x)=\frac{2}{x^{2/3}}\), \(f(x)=\frac{9}{2}x^{\frac{4}{3}}+C_1x + C_2\).
• For \(f''(x)=6x+\sin x\), \(f(x)=x^{3}-\sin x+C_1x + C_2\).