What is the integral from 0 to 1 of x^3(x^4-4)^6 ?

The integral from 0 to 1 of x^3(x^4-4)^6 is 14197/28

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integral from 0 to 1 of x^3(x^4-4)^6

\int_{0}^{1} x^{3} (x^{4} - 4)^{6} d x

\frac{14197}{28}

Decimal

507.03571 \dots

Solution steps

\int_{0}^{1} x^{3} (x^{4} - 4)^{6} d x

Apply u-substitution

: \int_{0}^{1} \frac{u ^{\frac{1}{3}} ( u ^{\frac{4}{3}} - 4 ) ^{6}}{3} d u

= \int_{0}^{1} \frac{u ^{\frac{1}{3}} ( u ^{\frac{4}{3}} - 4 ) ^{6}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} u^{\frac{1}{3}} (u^{\frac{4}{3}} - 4)^{6} d u

Expand

u^{\frac{1}{3}} (u^{\frac{4}{3}} - 4)^{6} : u^{\frac{25}{3}} - 24 u^{7} + 240 u^{\frac{17}{3}} - 1280 u^{\frac{13}{3}} + 3840 u^{3} - 6144 u^{\frac{5}{3}} + 4096 u^{\frac{1}{3}}

= \frac{1}{3} \cdot \int_{0}^{1} u^{\frac{25}{3}} - 24 u^{7} + 240 u^{\frac{17}{3}} - 1280 u^{\frac{13}{3}} + 3840 u^{3} - 6144 u^{\frac{5}{3}} + 4096 u^{\frac{1}{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int_{0}^{1} u^{\frac{25}{3}} d u - \int_{0}^{1} 24 u^{7} d u + \int_{0}^{1} 240 u^{\frac{17}{3}} d u - \int_{0}^{1} 1280 u^{\frac{13}{3}} d u + \int_{0}^{1} 3840 u^{3} d u - \int_{0}^{1} 6144 u^{\frac{5}{3}} d u + \int_{0}^{1} 4096 u^{\frac{1}{3}} d u)

\int_{0}^{1} u^{\frac{25}{3}} d u = \frac{3}{28}

\int_{0}^{1} 24 u^{7} d u = 3

\int_{0}^{1} 240 u^{\frac{17}{3}} d u = 36

\int_{0}^{1} 1280 u^{\frac{13}{3}} d u = 240

\int_{0}^{1} 3840 u^{3} d u = 960

\int_{0}^{1} 6144 u^{\frac{5}{3}} d u = 2304

\int_{0}^{1} 4096 u^{\frac{1}{3}} d u = 3072

= \frac{1}{3} (\frac{3}{28} - 3 + 36 - 240 + 960 - 2304 + 3072)

Simplify

\frac{1}{3} (\frac{3}{28} - 3 + 36 - 240 + 960 - 2304 + 3072) : \frac{14197}{28}

= \frac{14197}{28}

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What is the integral from 0 to 1 of x^3(x^4-4)^6 ?
The integral from 0 to 1 of x^3(x^4-4)^6 is 14197/28