Question

peer instruction: inclass_nov07
if ( f^{prime}(x)=9 x^{2}+sin (x)+2 ), and ( f(0)=3 ), then ( f(x) ) is given by
choose one
a ( f(x)=3 x^{3}+cos (x)+2 x+2 )
b ( f(x)=3 x^{3}-cos (x)+2 x+4 )
c ( f(x)=3 x^{3}+cos (x)+2 )
d ( f(x)=3 x^{3}-cos (x)+2 x+3 )
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activity apex-peer-antider-2 multiple choice (apex-peer-antider-2)
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Explanation:

Step1: Integrate \(f'(x)\)

Integrate each term of \(f'(x)=9x^{2}+\sin(x)+2\) separately.
Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), \(\int\sin(x)dx=-\cos(x)+C\), and \(\int kdx=kx + C\) (where \(k\) is a constant).
\(\int(9x^{2}+\sin(x)+2)dx=9\int x^{2}dx+\int\sin(x)dx+2\int dx\)
\(=9\times\frac{x^{3}}{3}-\cos(x)+2x + C\)
\(=3x^{3}-\cos(x)+2x + C\)

Step2: Use the initial condition \(f(0) = 3\)

Substitute \(x = 0\) and \(f(0)=3\) into \(f(x)=3x^{3}-\cos(x)+2x + C\).
\(f(0)=3\times0^{3}-\cos(0)+2\times0 + C\)
Since \(\cos(0)=1\), we have \(3=- 1+C\).
Solving for \(C\): \(C=3 + 1=4\)

Answer:

B. \(f(x)=3x^{3}-\cos(x)+2x + 4\)