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

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

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integral from-1 to 2 of sqrt(x)

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

\frac{4 2}{3} + i \frac{2}{3}

Solution steps

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

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{- 1}^{2}

= [\frac{2}{3} x^{\frac{3}{2}}]_{- 1}^{2}

Compute the boundaries:

\frac{4 2}{3} + i \frac{2}{3}

= \frac{4 2}{3} + i \frac{2}{3}

cos (x + \frac{π}{2})

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

\int t (t + 1)^{6} d t

f (x) = x^{2} - 1, at x = 2

y = e^{x}, y = e^{2 x}, x = 0, x = ln (2)

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