What is the integral from 1 to sqrt(x of)18t^7 ?

The integral from 1 to sqrt(x of)18t^7 is (9(x^4-1))/4

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integral from 1 to sqrt(x of)18t^7

\int_{1}^{x} 18 t^{7} d t

\frac{9 ( x ^{4} - 1 )}{4}

Solution steps

\int_{1}^{x} 18 t^{7} d t

Take the constant out:

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

= 18 \cdot \int_{1}^{x} t^{7} d t

Apply the Power Rule

: [\frac{t ^{8}}{8}]_{1}^{x}

= 18 [\frac{t ^{8}}{8}]_{1}^{x}

Compute the boundaries:

\frac{x ^{4}}{8} - \frac{1}{8}

= 18 (\frac{x ^{4}}{8} - \frac{1}{8})

Simplify

= \frac{9 ( x ^{4} - 1 )}{4}

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What is the integral from 1 to sqrt(x of)18t^7 ?
The integral from 1 to sqrt(x of)18t^7 is (9(x^4-1))/4