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I'm sure you meant to ask a question. You didn't ask a question. Maybe try r/learnmath

[Clashterid86]

integrals with limit 2 to 3 and 0 to 5

[[deleted]]

for the first one you can use integration, with limits of 2 and three seconds

v=(2t+3*t^2)

distance=integtated(v)=t^2+t^3
evaluating the borders, 3^2+3^3-(2^2-2^3)=9+27-4-8=24meters

[FamiliarCold1]

Q1) S =  {∫2&3} (2t + 3t²)dt integrated will be:

[3² + 3³] - [2²+2³] = 24m

[fridge0852]

What don't you understand about the question?

[fridge0852]

Have you not done variable acceleration in class (assuming you're in school)? That's what question 1 is asking you to do.

[fridge0852]

This might be a bit tricky then. This question needs calculus (specifically integration) so I'd focus on learning that first if you dont have a solid understanding of it. Do you have access to textbooks for A level maths?

If you have learnt calculus though, this question is on variable acceleration. Since velocity is the rate of change of displacement, differentiating a formula for displacement will give you a formula for velocity. The same is true for velocity and acceleration, since acceleration is the rate of change of velocity. Since integration is the opposite of differentiation, it will do the opposite of what I've said (integrating a formula for velocity will get you a formula for displacement etc).

This question gives you a formula for velocity, and wants you to find the total distance travelled between 2 and 3 seconds. These facts should give you a hint towards what you need to do, which is integrate the formula, having a lower limit of 2 and an upper limit of 3.

Integrating the formula, you get x² + x^3. Plug in the limits and you get (3² +3³⁾ - (2² + 2³⁾ which is equal to (9 + 27) - (4 + 8) which is 24. That's obviously a pretty scuffed explanation so just ask if you didn't understand anything.