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evaluate the integral (int x^2 (x^3 - 9)^{45} dx), by making the substi…

Question

evaluate the integral (int x^2 (x^3 - 9)^{45} dx), by making the substitution (u = x^3 - 9). (\boxed{} + c) note: your answer should be in terms of (x) and not (u).

Explanation:

Step1: Substitute \(u = x^3 - 9\)

Compute \(du/dx = 3x^2\), so \(du = 3x^2 dx\) → \(x^2 dx = \frac{1}{3}du\)

Step2: Rewrite integral in terms of \(u\)

Integral becomes \(\int u^{45} \cdot \frac{1}{3}du = \frac{1}{3}\int u^{45}du\)

Step3: Integrate with power rule

Power rule: \(\int u^n du = \frac{u^{n+1}}{n+1}\). Here \(n=45\), so \(\frac{1}{3} \cdot \frac{u^{46}}{46} + C = \frac{u^{46}}{138} + C\)

Step4: Substitute back \(u = x^3 -9\)

Result: \(\frac{(x^3 -9)^{46}}{138} + C\)

Answer:

\(\frac{(x^3 - 9)^{46}}{138}\)