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

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

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integral from 1 to x of (2-t)

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

2 x - 2 - \frac{x ^{2} - 1}{2}

Solution steps

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

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

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

= 2 x - 2 - \frac{x ^{2} - 1}{2}

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

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

\int_{0}^{\frac{π}{2}} (6 + 3 sin (x)) d x

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