What is the integral from-2 to x of (16t^2-15t) ?

The integral from-2 to x of (16t^2-15t) is (16(x^3+8))/3-15((x^2)/2-2)

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integral from-2 to x of (16t^2-15t)

\int_{- 2}^{x} (16 t^{2} - 15 t) d t

\frac{16 ( x ^{3} + 8 )}{3} - 15 (\frac{x ^{2}}{2} - 2)

Solution steps

\int_{- 2}^{x} 16 t^{2} - 15 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} 16 t^{2} d t - \int_{- 2}^{x} 15 t d t

\int_{- 2}^{x} 16 t^{2} d t = \frac{16 ( x ^{3} + 8 )}{3}

\int_{- 2}^{x} 15 t d t = 15 (\frac{x ^{2}}{2} - 2)

= \frac{16 ( x ^{3} + 8 )}{3} - 15 (\frac{x ^{2}}{2} - 2)

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What is the integral from-2 to x of (16t^2-15t) ?
The integral from-2 to x of (16t^2-15t) is (16(x^3+8))/3-15((x^2)/2-2)