the indefinite integral can be found in more than one way. first use the substitution method to find the…

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.\nuse the substitution method to find the indefinite integral. what expression should be used for the substitution, u?\na. 3x\nb. x³ - 3\nc. x\nd. x³\nfind the indefinite integral.\n∫3x²(x³ - 3)dx = □

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.\nuse the substitution method to find the indefinite integral. what expression should be used for the substitution, u?\na. 3x\nb. x³ - 3\nc. x\nd. x³\nfind the indefinite integral.\n∫3x²(x³ - 3)dx = □

Answer

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\neq- 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\neq-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)