What is the integral of 0.55t^2+1.575t^4 ?

The integral of 0.55t^2+1.575t^4 is 0.18333…t^3+0.315t^5+C

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integral of 0.55t^2+1.575t^4

\int 0.55 t^{2} + 1.575 t^{4} d t

0.18333 \dots t^{3} + 0.315 t^{5} + C

Solution steps

\int 0.55 t^{2} + 1.575 t^{4} 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 0.55 t^{2} d t + \int 1.575 t^{4} d t

\int 0.55 t^{2} d t = 0.18333 \dots t^{3}

\int 1.575 t^{4} d t = 0.315 t^{5}

= 0.18333 \dots t^{3} + 0.315 t^{5}

Add a constant to the solution

= 0.18333 \dots t^{3} + 0.315 t^{5} + C

\int \frac{x}{x ^{- \frac{5}{2}}} d x

\int 3 x^{2} - 3 y^{2} d y

\frac{d}{d t} (t e^{(t)})

x \to - 1 lim (4 x^{3} - 3 x^{2} + 5)

x \to 0 = lim (\frac{sin ( 4 x )}{6})

What is the integral of 0.55t^2+1.575t^4 ?
The integral of 0.55t^2+1.575t^4 is 0.18333…t^3+0.315t^5+C