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

The integral from-2 to 1 of (x+1)sqrt(x+3) is 46/15

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

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

\frac{46}{15}

Decimal Notation

3.06666 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} u (u - 2) d u

= \int_{1}^{4} u (u - 2) d u

Expand

u (u - 2) : u^{\frac{3}{2}} - 2 u

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

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

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} 2 u d u = \frac{28}{3}

= \frac{62}{5} - \frac{28}{3}

Simplify

\frac{62}{5} - \frac{28}{3} : \frac{46}{15}

= \frac{46}{15}

x^{3} + 2 x - 1

F (x) = 7 x^{6} (x^{4} - 7 x)

\frac{d}{d x} (tan (3 x))

f (x) = 2 x^{2}, at x = 4

\int cos (π x) d x

What is the integral from-2 to 1 of (x+1)sqrt(x+3) ?
The integral from-2 to 1 of (x+1)sqrt(x+3) is 46/15