What is the integral from 1 to 7 of (t^2-4t-12) ?

The integral from 1 to 7 of (t^2-4t-12) is -54

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integral from 1 to 7 of (t^2-4t-12)

\int_{1}^{7} (t^{2} - 4 t - 12) d t

- 54

Solution steps

\int_{1}^{7} t^{2} - 4 t - 12 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}^{7} t^{2} d t - \int_{1}^{7} 4 t d t - \int_{1}^{7} 12 d t

\int_{1}^{7} t^{2} d t = 114

\int_{1}^{7} 4 t d t = 96

\int_{1}^{7} 12 d t = 72

= 114 - 96 - 72

Simplify

= - 54

x \to 4 - lim (1 + 4 - x)

y = (x^{5} + 2)^{2} (x^{4} + 4)^{4}

n = 1 \sum \infty (- 1)^{n} sin (\frac{2}{n})

\frac{d}{d x} (6 x + 4)

\int_{1}^{3} 3 r^{4} ln (r) d r

What is the integral from 1 to 7 of (t^2-4t-12) ?
The integral from 1 to 7 of (t^2-4t-12) is -54