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

The integral from-1 to x^2 of (-2t+2) is -x^4+2x^2+3

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

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

- x^{4} + 2 x^{2} + 3

Solution steps

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

\int_{- 1}^{x^{2}} 2 t d t = x^{4} - 1

\int_{- 1}^{x^{2}} 2 d t = 2 x^{2} + 2

= - (x^{4} - 1) + 2 x^{2} + 2

Simplify

= - x^{4} + 2 x^{2} + 3

\frac{5}{x}

\frac{1}{e ^{y}}

y = \frac{- 2 y + x + 2}{2 y - x + 4}, (- 3, - 1)

f (x) = - sin (3 x) \cdot 3

\frac{\partial}{\partial z} (- yz)

What is the integral from-1 to x^2 of (-2t+2) ?
The integral from-1 to x^2 of (-2t+2) is -x^4+2x^2+3