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Popular >

integral from 0 to 1 of t+t^2

Solution

\int_{0}^{1} t + t^{2} d t

Solution

\frac{5}{6}

+1

Decimal

0.83333 \dots

Solution steps

\int_{0}^{1} t + t^{2} d t

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}^{1} t d t + \int_{0}^{1} t^{2} d t

\int_{0}^{1} t d t = \frac{1}{2}

\int_{0}^{1} t^{2} d t = \frac{1}{3}

= \frac{1}{2} + \frac{1}{3}

Simplify

\frac{1}{2} + \frac{1}{3} : \frac{5}{6}

= \frac{5}{6}

Graph

Popular Examples

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\int_{3}^{5} x^{2} - 8 x + 16 d x

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

integral from 0 to 1 of xe^{4x}

\int_{0}^{1} x e^{4 x} d x

integral from 0 to 1 of xsin((npix)/1)

\int_{0}^{1} x sin (\frac{nπ x}{1}) d x

integral from 0 to 6 of (2x-1)^2

\int_{0}^{6} (2 x - 1)^{2} d x

Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of t+t^2 ?
The integral from 0 to 1 of t+t^2 is 5/6