What is the integral from 0 to 1 of (3x+1)^4 ?

The integral from 0 to 1 of (3x+1)^4 is 341/5

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integral from 0 to 1 of (3x+1)^4

\int_{0}^{1} (3 x + 1)^{4} d x

\frac{341}{5}

Decimal Notation

68.2

Solution steps

\int_{0}^{1} (3 x + 1)^{4} d x

Apply u-substitution

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

= \int_{1}^{4} \frac{u ^{4}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{1}^{4} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{4}

= \frac{1}{3} [\frac{u ^{5}}{5}]_{1}^{4}

Compute the boundaries:

\frac{1023}{5}

= \frac{1}{3} \cdot \frac{1023}{5}

Simplify

= \frac{341}{5}

f (x) = \frac{1}{1 + x ^{2}}, at x = 2

\frac{d x}{d t} = x^{2} + \frac{1}{49}, x (0) = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 6

ln (x + 1)

\frac{d ^{3}}{d x ^{3}} (sin (7 x))

What is the integral from 0 to 1 of (3x+1)^4 ?
The integral from 0 to 1 of (3x+1)^4 is 341/5