What is the integral from 1 to 4 of sqrt(2x+1) ?

The integral from 1 to 4 of sqrt(2x+1) is 9-sqrt(3)

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integral from 1 to 4 of sqrt(2x+1)

\int_{1}^{4} 2 x + 1 d x

9 - 3

Decimal Notation

7.26794 \dots

Solution steps

\int_{1}^{4} 2 x + 1 d x

Apply u-substitution

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

= \int_{3}^{9} \frac{u}{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_{3}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{9}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{9}

Compute the boundaries:

18 - 23

= \frac{1}{2} (18 - 23)

Simplify

= 9 - 3

\int_{- 1}^{1} x^{2} 1 - x^{2} d x

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

\int_{0}^{\frac{π}{2}} cos (x) + sin (x) d x

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

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

What is the integral from 1 to 4 of sqrt(2x+1) ?
The integral from 1 to 4 of sqrt(2x+1) is 9-sqrt(3)