20. [-/1 Points] DETAILS SCALCET9M 13.2.037. Evaluate the integral. $\int_0^2 (4t \mathbf{i} - t^3 \mathbf{j} + 2t^3 \mathbf{k}) dt$ DETAILS SCALCET9M 13.2.040. 21. [-/1 Points] Evaluate the integral. $\int_0^{\pi/4} (\sec(t) \tan(t) \mathbf{i} + t \cos(2t) \mathbf{j} + \sin^2(2t) \cos(2t) \mathbf{k}) dt$ DETAILS SCALCET9M 13.2.041. 22. [-/1 Points] Evaluate the integral. (Use C for the constant of integration.) $\int \left( \frac{4}{1 + t^2} \mathbf{i} + te^t \mathbf{j} + 6 \sqrt{t} \mathbf{k} \right) dt$
Added by Mark M.
Close
Step 1
For the first component, we have 4t i. The integral of 4t with respect to t is 2t^2. So the integral of 4t i with respect to t is 2t^2 i. For the second component, we have -t^3 j. The integral of -t^3 with respect to t is -1/4 t^4. So the integral of -t^3 j with Show moreβ¦
Show all steps
Your feedback will help us improve your experience
Heba Qureshi and 74 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Evaluate the integral. $ \displaystyle \int_0^2 (t i - t^3 j + 3t^5 k) dt $
Vector Functions
Derivatives and Integrals of Vector Functions
Evaluate integral ? ?t/4t^2+1, t/3?t^2+1, t^2/(1+t^3) dt
Ahmet Y.
Evaluate the integral. $ \displaystyle \int_1^4 (2t^{\frac{3}{2}} i + (t + 1) \sqrt{t} k) dt $
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD