$int_{0}^{3}15w^{4}-13w^{2}+w dw$
$int_{0}^{3}15w^{4}-13w^{2}+w dw$
Answer
Explanation:
Step1: Apply power - rule for integration
The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int(15w^{4}-13w^{2}+w)dw=15\int w^{4}dw-13\int w^{2}dw+\int wdw$. $15\times\frac{w^{5}}{5}-13\times\frac{w^{3}}{3}+\frac{w^{2}}{2}=3w^{5}-\frac{13}{3}w^{3}+\frac{1}{2}w^{2}+C$.
Step2: Evaluate the definite integral
$\int_{0}^{3}(15w^{4}-13w^{2}+w)dw=\left[3w^{5}-\frac{13}{3}w^{3}+\frac{1}{2}w^{2}\right]_{0}^{3}$. First, substitute $w = 3$: $3\times3^{5}-\frac{13}{3}\times3^{3}+\frac{1}{2}\times3^{2}=3\times243-\frac{13}{3}\times27+\frac{1}{2}\times9$. $729 - 117+\frac{9}{2}$. $612+\frac{9}{2}=\frac{1224 + 9}{2}=\frac{1233}{2}$. Then substitute $w = 0$: $3\times0^{5}-\frac{13}{3}\times0^{3}+\frac{1}{2}\times0^{2}=0$. Subtract the two results: $\frac{1233}{2}-0=\frac{1233}{2}$.
Answer:
$\frac{1233}{2}$