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

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

\int (3) d x

f (x) = \frac{60}{0.01 x ^{2} + 1}

\frac{d}{d x} (arctan (x^{2} + 1))

n = 1 \sum \infty \frac{50}{5 ^{n}}

e^{l n (s e c (2 x))}

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