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Evaluate the integral ∫(2(6t i - t³ j + 5t⁹ k)) dt from 0

User Datajoely
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Final answer:

The integral of the vector function 2(6t i - t³ j + 5t⁹ k) from 0 to t is found by integrating each component separately, resulting in (6t² i - 0.5t⁴ j + 1t⁺ k).

Step-by-step explanation:

To evaluate the integral of a vector function a(t) = 2(6t i - t³ j + 5t⁹ k) from 0 to an arbitrary point t, we will integrate each component of the vector function separately.

Focusing on each component:

The integral of 2(6t) with respect to t over the interval from 0 to t for the i component gives us 6t² when evaluated from 0 to t, which simplifies to 6t².

The integral of -2t³ for the j component becomes -0.5t⁴, again evaluated from 0 to t, simplifying to -0.5t⁴.

The integral of 2(5t⁹) for the k component leads to 1t⁺ evaluated from 0 to t, which simplifies to 1t⁺.

Combining these results, the integrated vector function from 0 to t is (6t² i - 0.5t⁴ j + 1t⁺ k).

User Bhanu Sinha
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