Question

the indefinite integral can be found in more than one way. first use the substitution method to find the indefinite integral. then find it without using substitution. check that your answers are equivalent.
use the substitution method to find the indefinite integral. what expression should be used for the substitution, u?
a. 3x
b. x³ - 3
c. x
d. x³
find the indefinite integral.
∫3x²(x³ - 3)dx = □

Explanation:

Step1: Substitute \(u = x^{3}-3\)

Differentiate \(u\) with respect to \(x\): \(du=3x^{2}dx\).

Step2: Rewrite the integral

The integral \(\int3x^{2}(x^{3} - 3)dx\) becomes \(\int udu\) (since \(3x^{2}dx=du\) and \(x^{3}-3 = u\)).

Step3: Integrate \(\int udu\)

Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq- 1\)), we have \(\int udu=\frac{u^{2}}{2}+C\).

Step4: Substitute back \(u = x^{3}-3\)

We get \(\frac{(x^{3}-3)^{2}}{2}+C=\frac{x^{6}-6x^{3}+9}{2}+C=\frac{1}{2}x^{6}-3x^{3}+\frac{9}{2}+C\).

Another way (expanding the integrand first):

Step1: Expand \(3x^{2}(x^{3}-3)\)

\(3x^{2}(x^{3}-3)=3x^{5}-9x^{2}\).

Step2: Integrate term - by - term

\(\int(3x^{5}-9x^{2})dx\). Using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq-1\)), \(\int3x^{5}dx=3\times\frac{x^{6}}{6}=\frac{1}{2}x^{6}\) and \(\int(-9x^{2})dx=-9\times\frac{x^{3}}{3}=-3x^{3}\). So \(\int(3x^{5}-9x^{2})dx=\frac{1}{2}x^{6}-3x^{3}+C\).

Answer:

\(\frac{1}{2}x^{6}-3x^{3}+C\)