What is the integral from 2x^2 to x^4 of 2t+t^2 ?

The integral from 2x^2 to x^4 of 2t+t^2 is x^8-4x^4+(x^{12}-8x^6)/3

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integral from 2x^2 to x^4 of 2t+t^2

\int_{2 x^{2}}^{x^{4}} 2 t + t^{2} d t

x^{8} - 4 x^{4} + \frac{x ^{12} - 8 x ^{6}}{3}

Solution steps

\int_{2 x^{2}}^{x^{4}} 2 t + t^{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_{2 x^{2}}^{x^{4}} 2 t d t + \int_{2 x^{2}}^{x^{4}} t^{2} d t

\int_{2 x^{2}}^{x^{4}} 2 t d t = x^{8} - 4 x^{4}

\int_{2 x^{2}}^{x^{4}} t^{2} d t = \frac{x ^{12} - 8 x ^{6}}{3}

= x^{8} - 4 x^{4} + \frac{x ^{12} - 8 x ^{6}}{3}

\int_{- 2}^{0} (3 x 4 - x^{2}) d x

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\int_{- 2}^{0} - c x d x

\int_{0}^{π} e^{3 x} \cdot sin (x) d x

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

What is the integral from 2x^2 to x^4 of 2t+t^2 ?
The integral from 2x^2 to x^4 of 2t+t^2 is x^8-4x^4+(x^{12}-8x^6)/3