What is the integral from-4 to x of (t-1) ?

The integral from-4 to x of (t-1) is (x^2)/2-x-12

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integral from-4 to x of (t-1)

\int_{- 4}^{x} (t - 1) d t

\frac{x ^{2}}{2} - x - 12

Solution steps

\int_{- 4}^{x} t - 1 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_{- 4}^{x} t d t - \int_{- 4}^{x} 1 d t

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

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

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

Simplify

= \frac{x ^{2}}{2} - x - 12

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What is the integral from-4 to x of (t-1) ?
The integral from-4 to x of (t-1) is (x^2)/2-x-12