What is the integral from 1 to 3x of (t^2+t) ?

The integral from 1 to 3x of (t^2+t) is 9x^3-1/3+(9x^2-1)/2

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integral from 1 to 3x of (t^2+t)

\int_{1}^{3 x} (t^{2} + t) d t

9 x^{3} - \frac{1}{3} + \frac{9 x ^{2} - 1}{2}

Solution steps

\int_{1}^{3 x} t^{2} + t 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_{1}^{3 x} t^{2} d t + \int_{1}^{3 x} t d t

\int_{1}^{3 x} t^{2} d t = 9 x^{3} - \frac{1}{3}

\int_{1}^{3 x} t d t = \frac{9 x ^{2} - 1}{2}

= 9 x^{3} - \frac{1}{3} + \frac{9 x ^{2} - 1}{2}

f (x) = 2 - 2 x^{2}, - 16, at x = 3

x \to - 4 lim (10 x^{2} + 34 x - 24)

n = 1 \sum \infty 2 - 5 (0.9)^{n}

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

x \to - 5 - lim (\frac{x + 2}{x + 5})

What is the integral from 1 to 3x of (t^2+t) ?
The integral from 1 to 3x of (t^2+t) is 9x^3-1/3+(9x^2-1)/2