What is the integral of 24t-t^2+5t^{1/2} ?

The integral of 24t-t^2+5t^{1/2} is 12t^2-(t^3)/3+10/3 t^{3/2}+C

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integral of 24t-t^2+5t^{1/2}

\int 24 t - t^{2} + 5 t^{\frac{1}{2}} d t

12 t^{2} - \frac{t ^{3}}{3} + \frac{10}{3} t^{\frac{3}{2}} + C

Solution steps

\int 24 t - t^{2} + 5 t^{\frac{1}{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 24 t d t - \int t^{2} d t + \int 5 t^{\frac{1}{2}} d t

\int 24 t d t = 12 t^{2}

\int t^{2} d t = \frac{t ^{3}}{3}

\int 5 t^{\frac{1}{2}} d t = \frac{10}{3} t^{\frac{3}{2}}

= 12 t^{2} - \frac{t ^{3}}{3} + \frac{10}{3} t^{\frac{3}{2}}

Add a constant to the solution

= 12 t^{2} - \frac{t ^{3}}{3} + \frac{10}{3} t^{\frac{3}{2}} + C

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What is the integral of 24t-t^2+5t^{1/2} ?
The integral of 24t-t^2+5t^{1/2} is 12t^2-(t^3)/3+10/3 t^{3/2}+C