What is the integral from 0 to 2 of 8t ?

The integral from 0 to 2 of 8t is 16

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integral from 0 to 2 of 8t

\int_{0}^{2} 8 t d t

16

Solution steps

\int_{0}^{2} 8 t d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{2} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{2}

= 8 [\frac{t ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= 8 \cdot 2

Simplify

= 16

f (t) = t

\frac{d}{d x} (- \frac{2 x}{( x ^{2} - 4 ) ^{2}})

\int \frac{12}{x ^{2} + 1} d x

(1 - x)^{\frac{1}{5}}