What is the integral from-5 to 0 of 2(t+1)^{1/3} ?

The integral from-5 to 0 of 2(t+1)^{1/3} is 2(3/4-3*4^{1/3})

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integral from-5 to 0 of 2(t+1)^{1/3}

\int_{- 5}^{0} 2 (t + 1)^{\frac{1}{3}} d t

2 (\frac{3}{4} - 3 \cdot 4^{\frac{1}{3}})

Decimal Notation

- 8.02440 \dots

Solution steps

\int_{- 5}^{0} 2 (t + 1)^{\frac{1}{3}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 5}^{0} (t + 1)^{\frac{1}{3}} d t

Apply u-substitution

: \int_{- 4}^{1} u^{\frac{1}{3}} d u

= 2 \cdot \int_{- 4}^{1} u^{\frac{1}{3}} d u

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{- 4}^{1}

= 2 [\frac{3}{4} u^{\frac{4}{3}}]_{- 4}^{1}

Compute the boundaries:

\frac{3}{4} - 3 \cdot 4^{\frac{1}{3}}

= 2 (\frac{3}{4} - 3 \cdot 4^{\frac{1}{3}})

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

\int_{0}^{2} x e^{5 x} d x

\int_{- 10}^{10} (∣ x ∣ + 5) d x

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

What is the integral from-5 to 0 of 2(t+1)^{1/3} ?
The integral from-5 to 0 of 2(t+1)^{1/3} is 2(3/4-3*4^{1/3})