What is the integral from-1 to 1 of 7t^6 ?

The integral from-1 to 1 of 7t^6 is 2

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integral from-1 to 1 of 7t^6

\int_{- 1}^{1} 7 t^{6} d t

2

Solution steps

\int_{- 1}^{1} 7 t^{6} d t

Take the constant out:

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

= 7 \cdot \int_{- 1}^{1} t^{6} d t

Apply the Power Rule

: [\frac{t ^{7}}{7}]_{- 1}^{1}

= 7 [\frac{t ^{7}}{7}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{7}

= 7 \cdot \frac{2}{7}

Simplify

= 2

\int_{0}^{1} t^{2} + 2 d t

\int_{y}^{y} 1 d x

\int_{0}^{2} \frac{1}{6} x d x

\int_{1}^{4} \frac{1}{6} (x - 1) d x

\int_{25}^{40} \frac{1}{15} x^{2} d x

What is the integral from-1 to 1 of 7t^6 ?
The integral from-1 to 1 of 7t^6 is 2