What is the integral from-2 to 2 of (3ti+2t^2j-t^3k) ?

The integral from-2 to 2 of (3ti+2t^2j-t^3k) is (32j)/3

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integral from-2 to 2 of (3ti+2t^2j-t^3k)

\int_{- 2}^{2} (3 t i + 2 t^{2} j - t^{3} k) d t

\frac{32 j}{3}

Solution steps

\int_{- 2}^{2} 3 t i + 2 t^{2} j - t^{3} k 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_{- 2}^{2} 3 t i d t + \int_{- 2}^{2} 2 t^{2} j d t - \int_{- 2}^{2} t^{3} k d t

\int_{- 2}^{2} 3 t i d t = 0

\int_{- 2}^{2} 2 t^{2} j d t = \frac{32 j}{3}

\int_{- 2}^{2} t^{3} k d t = 0

= 0 + \frac{32 j}{3} - 0

Simplify

= \frac{32 j}{3}

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

\int_{0}^{y} x^{2} d x

\int_{1}^{4} (1 + \frac{1}{u} + \frac{1}{u ^{2}}) d u

\int_{1}^{0} x^{3} d x

\int_{0}^{2} 2 x x^{2} - 3 x + 2 d x

What is the integral from-2 to 2 of (3ti+2t^2j-t^3k) ?
The integral from-2 to 2 of (3ti+2t^2j-t^3k) is (32j)/3