What is the integral from-2 to x of (4t^2-3t) ?

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

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integral from-2 to x of (4t^2-3t)

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

\frac{4 ( x ^{3} + 8 )}{3} - 3 (\frac{x ^{2}}{2} - 2)

Solution steps

\int_{- 2}^{x} 4 t^{2} - 3 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_{- 2}^{x} 4 t^{2} d t - \int_{- 2}^{x} 3 t d t

\int_{- 2}^{x} 4 t^{2} d t = \frac{4 ( x ^{3} + 8 )}{3}

\int_{- 2}^{x} 3 t d t = 3 (\frac{x ^{2}}{2} - 2)

= \frac{4 ( x ^{3} + 8 )}{3} - 3 (\frac{x ^{2}}{2} - 2)

\int_{0}^{1} \frac{1}{x ^{3} + x ^{2} + 2 x + 2} d x

\int_{0}^{\frac{π}{2}} 2 π x cos (x) d x

\int_{3}^{6} x^{2} - 3 x d x

\int_{0}^{3} 6 x - x^{2} d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) - 1 d x

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