What is the integral from x^4 to x^6 of (2t-1)^3 ?

The integral from x^4 to x^6 of (2t-1)^3 is x^4(2x^{20}-4x^{14}-2x^{12}+7x^8-3x^4-x^2+1)

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integral from x^4 to x^6 of (2t-1)^3

\int_{x^{4}}^{x^{6}} (2 t - 1)^{3} d t

x^{4} (2 x^{20} - 4 x^{14} - 2 x^{12} + 7 x^{8} - 3 x^{4} - x^{2} + 1)

Solution steps

\int_{x^{4}}^{x^{6}} (2 t - 1)^{3} d t

Apply u-substitution

: \int_{2 x^{4} - 1}^{2 x^{6} - 1} \frac{u ^{3}}{2} d u

= \int_{2 x^{4} - 1}^{2 x^{6} - 1} \frac{u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2 x^{4} - 1}^{2 x^{6} - 1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2 x^{4} - 1}^{2 x^{6} - 1}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{2 x^{4} - 1}^{2 x^{6} - 1}

Compute the boundaries:

\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{4} - 1 ) ^{4}}{4}

= \frac{1}{2} (\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{4} - 1 ) ^{4}}{4})

Simplify

= x^{4} (2 x^{20} - 4 x^{14} - 2 x^{12} + 7 x^{8} - 3 x^{4} - x^{2} + 1)

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What is the integral from x^4 to x^6 of (2t-1)^3 ?
The integral from x^4 to x^6 of (2t-1)^3 is x^4(2x^{20}-4x^{14}-2x^{12}+7x^8-3x^4-x^2+1)