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

The integral from x^4 to x^3 of (2t-1)^3 is x^3(-2x^{13}+6x^9-4x^6-3x^5+3x^3+x-1)

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

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

x^{3} (- 2 x^{13} + 6 x^{9} - 4 x^{6} - 3 x^{5} + 3 x^{3} + x - 1)

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

= x^{3} (- 2 x^{13} + 6 x^{9} - 4 x^{6} - 3 x^{5} + 3 x^{3} + x - 1)

\int_{0}^{\infty} 2 y e^{- 2 y} d y

\int_{0}^{x} \frac{t - 3}{t ^{2} + 7} d t

\int_{3}^{6} x^{2} - 2 x d x

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

\int_{0}^{7} x e^{x} d x

What is the integral from x^4 to x^3 of (2t-1)^3 ?
The integral from x^4 to x^3 of (2t-1)^3 is x^3(-2x^{13}+6x^9-4x^6-3x^5+3x^3+x-1)