Integral Proble, Please help asap, thank you very much.

[Integral]

Integral Problem, Please help asap

Hello, I have been trying to do this for days and I really can't do it.Please help me.Problem:1) a) Calculate the area of ​​the area bounded by y = 3x ^ 3 +2 x +1 and the x-axis between x = 2 and x = 5. b) https://i.minus.com/iRLPzhGy1vZ5E.PNG Use new integration variable when changing terminals.(in french: Utiliser de nouvelles bornes d'intégration lors du changement de variable. )Thank you

 

[JeffM]

Hello, I have been trying to do this for days and I really can't do it.Please help me.Problem:1) a) Calculate the area of ​​the area bounded by y = 3x ^ 3 +2 x +1 and the x-axis between x = 2 and x = 5. b) https://i.minus.com/iRLPzhGy1vZ5E.PNG Use new integration variable when changing terminals.(in french: Utiliser de nouvelles bornes d'intégration lors du changement de variable. )Thank you

"Terminal" has peculiar meanings in English and is not appropriate in this context. The word "bornes" should be translated as "bounds" to be idiomatic English in this context. A better translation would be "Use new bounds of integration when changing the variable." Probablt denis will tell me to go to the corner because my French is long forgotten.

Please show us your work so we can see where you are going wrong. Are you simply asking how to calculate:

\(\displaystyle \displaystyle \int_2^5(3x^3 + 2x + 1)dx\)

By the way, please read Read Before Posting. Show your work, and do not post multiple problems in one thread.

 

[Integral]

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I really don't understand part b)
No idea how to do it.

a and b is basically one problem. I just separated them.

 

[JeffM]

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I really don't understand part b)
No idea how to do it.

a and b is basically one problem. I just separated them.

I'll get you started, but I have to leave. If you are still stuck, someone else will wander by to help.

\(\displaystyle x = 4t - \dfrac{\pi }{2} \implies \dfrac{dx}{dt} = 4 \implies dt = \dfrac{1}{4}dx.\)

\(\displaystyle And\ x = 4t - \dfrac{\pi }{2} \implies sin\left(4t - \dfrac{\pi }{2}\right) = sin(x)\ and\ cos\left(4t - \dfrac{\pi }{2}\right) = cos(x).\)

What next?

 

[Integral]

I'll get you started, but I have to leave. If you are still stuck, someone else will wander by to help.

\(\displaystyle x = 4t - \dfrac{\pi }{2} \implies \dfrac{dx}{dt} = 4 \implies dt = \dfrac{1}{4}dx.\)

\(\displaystyle And\ x = 4t - \dfrac{\pi }{2} \implies sin\left(4t - \dfrac{\pi }{2}\right) = sin(x)\ and\ cos\left(4t - \dfrac{\pi }{2}\right) = cos(x).\)

What next?

Can someone please continue it,
I really don't understand and I am so confused

 

[DrPhil]

I'll get you started, but I have to leave. If you are still stuck, someone else will wander by to help.

\(\displaystyle x = 4t - \dfrac{\pi }{2} \implies \dfrac{dx}{dt} = 4 \implies dt = \dfrac{1}{4}dx.\)

\(\displaystyle And\ x = 4t - \dfrac{\pi }{2} \implies sin\left(4t - \dfrac{\pi }{2}\right) = sin(x)\ and\ cos\left(4t - \dfrac{\pi }{2}\right) = cos(x).\)

What next?

What is next is to find the new limits of the integration, in terms of x instead of t:

\(\displaystyle t = 0 \implies x = -\pi/2\)

\(\displaystyle t = \dfrac {3 \pi}{8} \implies x = \pi \)

The integral becomes

\(\displaystyle \displaystyle \int_{-\pi/2}^{\pi}6\ \sin{x}\ \cos{x}\ dx\)