The indefinite integral of ⟨1,2t,et⟩ is a)⟨0,2,et⟩ b)⟨t,t2,et⟩+C. c)⟨t+C,t2+C,et+c⟩.

Question

The indefinite integral of ⟨1,2t,et⟩ is  a)⟨0,2,et⟩  b)⟨t,t2,et⟩+C.  c)⟨t+C,t2+C,et+c⟩.

Thomas Conway · Accepted Answer

Step 1 A function of the form r(t)=f(t)i^+g(t)j^+h(t)k^orr(t)=⟨f(t),g(t),h(t)⟩ is called a vector-valued function in 3-D space, where f(t), g(t), and h(t) are the component functions depending on the parameter t If R(t) is an anti-derivative of r(t), the indefinite integral of r(t) in component form is given by R(t)=∫r(t)dt =∫dt =&amp;lt;∫f(t)dt,∫g(t)dt,∫h(t)dt&amp;gt;dt Step 2 We have the given function as r(t)=&amp;lt;1,2t,et&amp;gt; Now, indefinite integral of r(t) in component form is given by ∫r(t)dt=∫&amp;lt;1,2t,et&amp;gt;dt =&amp;lt;∫{{(1)dt=∫{{(2t)dt,∫{{(et)dt&amp;gt; =&amp;lt;t+C,t2+C,et+C&amp;gt; Hence, option (c) is correct option.

The indefinite integral of <<1,2t,e^{t}>> is a) <<0,2,e^{t}>> b) <<

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