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

The integral from-1 to 6 of (x+3)^3 is 6545/4

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

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

\frac{6545}{4}

Decimal Notation

1636.25

Solution steps

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

Apply u-substitution

: \int_{2}^{9} u^{3} d u

= \int_{2}^{9} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{9}

= [\frac{u ^{4}}{4}]_{2}^{9}

Compute the boundaries:

\frac{6545}{4}

= \frac{6545}{4}

sin (x), cos (x), [\frac{π}{4}, \frac{5 π}{4}]

x^{2} y^{^{''}} - 3 x y^{^{'}} + 13 y = 0

\int arccot (9 y) d y

\int \frac{12 x ^{5}}{3 + 2 x ^{6}} d x

\frac{d}{d x} (sin^{4} (2 x))

What is the integral from-1 to 6 of (x+3)^3 ?
The integral from-1 to 6 of (x+3)^3 is 6545/4