What is area sqrt(x+2), 1/(x+1),[0,2] ?

The answer to area sqrt(x+2), 1/(x+1),[0,2] is (16-4sqrt(2))/3-ln(3)

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area sqrt(x+2), 1/(x+1),[0,2]

area

x + 2, \frac{1}{x + 1}, [0, 2]

\frac{16 - 4 2}{3} - ln (3)

Decimal Notation

2.34910 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} x + 2 - \frac{1}{x + 1} d x

Solve

\int_{0}^{2} x + 2 - \frac{1}{x + 1} d x : \frac{16 - 4 2}{3} - ln (3)

The area is:

= \frac{16 - 4 2}{3} - ln (3)

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n = 1 \sum \infty \frac{e ^{\frac{7}{n}}}{n}

What is area sqrt(x+2), 1/(x+1),[0,2] ?
The answer to area sqrt(x+2), 1/(x+1),[0,2] is (16-4sqrt(2))/3-ln(3)