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Popular >

integral from 0 to 5 of (x+1)

Solution

\int_{0}^{5} (x + 1) d x

Solution

\frac{35}{2}

+1

Decimal Notation

17.5

Solution steps

\int_{0}^{5} x + 1 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_{0}^{5} x d x + \int_{0}^{5} 1 d x

\int_{0}^{5} x d x = \frac{25}{2}

\int_{0}^{5} 1 d x = 5

= \frac{25}{2} + 5

Simplify

\frac{25}{2} + 5 : \frac{35}{2}

= \frac{35}{2}

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\int (θ^{2} - csc^{2} (θ)) d θ

area y=x^2,y=sqrt(7+x)area

y = x^{2}, y = 7 + x

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\int e^{4 x} sin (4 x) d x

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\frac{1}{( 1 - s ) \cdot ( s ^{2} + 1 )}

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y^{^{'}} (x) = \frac{9 x ^{2}}{9 y ^{2} - 7}, y (1) = 0

Frequently Asked Questions (FAQ)

What is the integral from 0 to 5 of (x+1) ?
The integral from 0 to 5 of (x+1) is 35/2