What is the integral from x^5 to x^6 of (2t-1)^3 ?

The integral from x^5 to x^6 of (2t-1)^3 is x^5(2x^{19}-2x^{15}-4x^{13}+4x^{10}+3x^7-3x^5-x+1)

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integral from x^5 to x^6 of (2t-1)^3

\int_{x^{5}}^{x^{6}} (2 t - 1)^{3} d t

x^{5} (2 x^{19} - 2 x^{15} - 4 x^{13} + 4 x^{10} + 3 x^{7} - 3 x^{5} - x + 1)

Solution steps

\int_{x^{5}}^{x^{6}} (2 t - 1)^{3} d t

Apply u-substitution

: \int_{2 x^{5} - 1}^{2 x^{6} - 1} \frac{u ^{3}}{2} d u

= \int_{2 x^{5} - 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^{5} - 1}^{2 x^{6} - 1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2 x^{5} - 1}^{2 x^{6} - 1}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{2 x^{5} - 1}^{2 x^{6} - 1}

Compute the boundaries:

\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{5} - 1 ) ^{4}}{4}

= \frac{1}{2} (\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{5} - 1 ) ^{4}}{4})

Simplify

= x^{5} (2 x^{19} - 2 x^{15} - 4 x^{13} + 4 x^{10} + 3 x^{7} - 3 x^{5} - x + 1)

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What is the integral from x^5 to x^6 of (2t-1)^3 ?
The integral from x^5 to x^6 of (2t-1)^3 is x^5(2x^{19}-2x^{15}-4x^{13}+4x^{10}+3x^7-3x^5-x+1)