Question

question 7
find the indefinite integral.
$\int(4 - 6x^{4}+2x^{8})dx$
$\frac{2}{9}x^{8}-\frac{6}{5}x^{4}+4x + c$
$\frac{2}{9}x^{9}-\frac{6}{5}x^{5}+4x + c$
no correct answer choice is given.
$4x - 24x^{3}+16x^{7}+c$
$4x - 6x^{5}+2x^{9}+c$

Explanation:

Step1: Apply the sum - rule of integration

The sum - rule states that \(\int(f(x)+g(x)+h(x))dx=\int f(x)dx+\int g(x)dx+\int h(x)dx\). So, \(\int(4 - 6x^{4}+2x^{8})dx=\int4dx-\int6x^{4}dx+\int2x^{8}dx\).

Step2: Use the power - rule of integration

The power - rule is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int kdx=kx + C\) (\(k\) is a constant).

• For \(\int4dx\): Using \(\int kdx=kx + C\) with \(k = 4\), we get \(4x+C_1\).
• For \(\int6x^{4}dx\): Using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) with \(n = 4\) and \(k = 6\), we have \(6\times\frac{x^{4+1}}{4 + 1}=\frac{6x^{5}}{5}+C_2\).
• For \(\int2x^{8}dx\): Using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) with \(n = 8\) and \(k = 2\), we get \(2\times\frac{x^{8+1}}{8+1}=\frac{2x^{9}}{9}+C_3\).

Step3: Combine the results

\(\int(4 - 6x^{4}+2x^{8})dx=4x-\frac{6x^{5}}{5}+\frac{2x^{9}}{9}+C\) (where \(C = C_1+C_2+C_3\)).

Answer:

\(\frac{2}{9}x^{9}-\frac{6}{5}x^{5}+4x + C\)