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

The integral from x^3 to x^6 of (2t-1)^3 is x^3(2x^{21}-4x^{15}+x^9+4x^6-4x^3+1)

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

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

x^{3} (2 x^{21} - 4 x^{15} + x^{9} + 4 x^{6} - 4 x^{3} + 1)

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2 x^{3} - 1}^{2 x^{6} - 1}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{2 x^{3} - 1}^{2 x^{6} - 1}

Compute the boundaries:

\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{3} - 1 ) ^{4}}{4}

= \frac{1}{2} (\frac{( 2 x ^{6} - 1 ) ^{4}}{4} - \frac{( 2 x ^{3} - 1 ) ^{4}}{4})

Simplify

= x^{3} (2 x^{21} - 4 x^{15} + x^{9} + 4 x^{6} - 4 x^{3} + 1)

\int_{1}^{2} x (x^{2} - 1)^{2} d x

\int_{2}^{4} 64 - x^{3} d x

\int_{2}^{5} (2 + 2 x) d x

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

\int_{- \frac{3}{2}}^{1} (x^{3} - 3 x) d x

What is the integral from x^3 to x^6 of (2t-1)^3 ?
The integral from x^3 to x^6 of (2t-1)^3 is x^3(2x^{21}-4x^{15}+x^9+4x^6-4x^3+1)