What is the integral of t^{5/2}(t^2-10t+16) ?

The integral of t^{5/2}(t^2-10t+16) is 2/11 t^{11/2}-20/9 t^{9/2}+32/7 t^{7/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of t^{5/2}(t^2-10t+16)

\int t^{\frac{5}{2}} (t^{2} - 10 t + 16) d t

\frac{2}{11} t^{\frac{11}{2}} - \frac{20}{9} t^{\frac{9}{2}} + \frac{32}{7} t^{\frac{7}{2}} + C

Solution steps

\int t^{\frac{5}{2}} (t^{2} - 10 t + 16) d t

Expand

t^{\frac{5}{2}} (t^{2} - 10 t + 16) : t^{\frac{9}{2}} - 10 t^{\frac{7}{2}} + 16 t^{\frac{5}{2}}

= \int t^{\frac{9}{2}} - 10 t^{\frac{7}{2}} + 16 t^{\frac{5}{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{\frac{9}{2}} d t - \int 10 t^{\frac{7}{2}} d t + \int 16 t^{\frac{5}{2}} d t

\int t^{\frac{9}{2}} d t = \frac{2}{11} t^{\frac{11}{2}}

\int 10 t^{\frac{7}{2}} d t = \frac{20}{9} t^{\frac{9}{2}}

\int 16 t^{\frac{5}{2}} d t = \frac{32}{7} t^{\frac{7}{2}}

= \frac{2}{11} t^{\frac{11}{2}} - \frac{20}{9} t^{\frac{9}{2}} + \frac{32}{7} t^{\frac{7}{2}}

Add a constant to the solution

= \frac{2}{11} t^{\frac{11}{2}} - \frac{20}{9} t^{\frac{9}{2}} + \frac{32}{7} t^{\frac{7}{2}} + C

What is the integral of t^{5/2}(t^2-10t+16) ?
The integral of t^{5/2}(t^2-10t+16) is 2/11 t^{11/2}-20/9 t^{9/2}+32/7 t^{7/2}+C