What is the integral from 0 to 1 of 3(1+x)^7 ?

The integral from 0 to 1 of 3(1+x)^7 is 765/8

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integral from 0 to 1 of 3(1+x)^7

\int_{0}^{1} 3 (1 + x)^{7} d x

\frac{765}{8}

Decimal Notation

95.625

Solution steps

\int_{0}^{1} 3 (1 + x)^{7} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} (1 + x)^{7} d x

Apply u-substitution

: \int_{1}^{2} u^{7} d u

= 3 \cdot \int_{1}^{2} u^{7} d u

Apply the Power Rule

: [\frac{u ^{8}}{8}]_{1}^{2}

= 3 [\frac{u ^{8}}{8}]_{1}^{2}

Compute the boundaries:

\frac{255}{8}

= 3 \cdot \frac{255}{8}

Simplify

= \frac{765}{8}

n = 1 \sum \infty 2 (0.8)^{n - 1}

ln (x^{3})

\frac{d}{d x} (a x^{3} + b x^{2} - \frac{c}{3} x + d)

y = x^{2} (x - 1)

\frac{\partial}{\partial x} (x^{7} y^{9} - x^{6} y^{8})

What is the integral from 0 to 1 of 3(1+x)^7 ?
The integral from 0 to 1 of 3(1+x)^7 is 765/8