Question

find the indefinite integral and check the result by differentiating.
int x ^ { 2 } ( 5 x + 4 ) d x
int x ^ { 2 } ( 5 x + 4 ) d x = (use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Expand the integrand

First, expand \(x^{2}(5x + 4)\) using the distributive property \(a(b + c)=ab+ac\). So, \(x^{2}(5x + 4)=5x^{3}+4x^{2}\). Then the integral becomes \(\int(5x^{3}+4x^{2})dx\).

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int5x^{3}dx\), since \(\int kf(x)dx=k\int f(x)dx\) (\(k = 5\) and \(f(x)=x^{3}\)), we have \(5\times\frac{x^{3+1}}{3 + 1}=\frac{5x^{4}}{4}\).
For \(\int4x^{2}dx\), using \(k = 4\) and \(f(x)=x^{2}\), we get \(4\times\frac{x^{2+1}}{2+1}=\frac{4x^{3}}{3}\).
So, \(\int(5x^{3}+4x^{2})dx=\frac{5x^{4}}{4}+\frac{4x^{3}}{3}+C\).

Step3: Check by differentiating

Differentiate \(y=\frac{5x^{4}}{4}+\frac{4x^{3}}{3}+C\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(y^\prime=\frac{5\times4x^{4-1}}{4}+\frac{4\times3x^{3 - 1}}{3}+0\).
Simplify \(y^\prime = 5x^{3}+4x^{2}\), which is the original integrand \(x^{2}(5x + 4)\) (after expanding).

Answer:

\(\frac{5x^{4}}{4}+\frac{4x^{3}}{3}+C\)