What is the integral from-1 to 0 of (2x+3)^2 ?

The integral from-1 to 0 of (2x+3)^2 is 13/3

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integral from-1 to 0 of (2x+3)^2

\int_{- 1}^{0} (2 x + 3)^{2} d x

\frac{13}{3}

Decimal

4.33333 \dots

Solution steps

\int_{- 1}^{0} (2 x + 3)^{2} d x

Expand

(2 x + 3)^{2} : 4 x^{2} + 12 x + 9

= \int_{- 1}^{0} 4 x^{2} + 12 x + 9 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{0} 4 x^{2} d x + \int_{- 1}^{0} 12 x d x + \int_{- 1}^{0} 9 d x

\int_{- 1}^{0} 4 x^{2} d x = \frac{4}{3}

\int_{- 1}^{0} 12 x d x = - 6

\int_{- 1}^{0} 9 d x = 9

= \frac{4}{3} - 6 + 9

Simplify

\frac{4}{3} - 6 + 9 : \frac{13}{3}

= \frac{13}{3}

\int e^{\frac{1}{x}} \frac{1}{x ^{3}} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = t^{2}

\frac{d}{d x} (\frac{e ^{- x} - x e ^{- x}}{e ^{x}})

e^{- 0.75 t} \cdot cos (4 t)

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

What is the integral from-1 to 0 of (2x+3)^2 ?
The integral from-1 to 0 of (2x+3)^2 is 13/3