# Having a problem in steps while solving integrals

• prakhargupta3301
In summary, the problem is that the incorrect factor (t+C) is used in the calculation of displacement in time 't' using integration. The correct solution is to use the integration of 3t+2 with respect to t.

#### prakhargupta3301

Member advised (again!) to use the formatting template for all homework help requests
Okay. So I'm new to calculus. And this is the first time I'm solving a physics problem using integration.
I understood that ∫dt (or ∫1dt) will be equal to just 't+C.' (Just like f ' (t+C) = 1).
Though, that's not the problem. The problem is when I apply it this way:

Question: The expression for velocity of a body is given as 3t+2. Displacement (x) is in metres and time (t) in seconds. Find the displacement (x) in time 't' using integration. (Note: take starting time to be 0 seconds)

My attempt: x= t0∫vdt
x = t0∫(3t+2)dt
x = [(3t2)/2 + 2t] ⋅ (t+C) ('t+C' is the integral of dt)
x = [(3t3)/2 +2t2] ⋅ (t+C)
x = [(3t2C)/2 + (3t3)/2]+ [(2tC+2t2)] t0|
x= [(3t2C)/2 + (3t3)/2]- 0
x [(3t2C)/2 + (3t3)/2] metres

However, as is evident (even to me) that this approach + method is wrong. The correct solution is:
x= t0∫vdt
x = t0∫(3t+2)dt
x = (3t2)/2 + 2t t0|
x= [(3t2)/2 + 2t]- 0
x= (3t2)/2 + 2t metres

Please tell me, why in step 3 of correct solution, integral of dt was omitted? (Please refer to same step of my wrong calculation to know the error in my approach)

Also, please enlighten me on the correct way.

prakhargupta3301 said:
why in step 3 of correct solution, integral of dt was omitted?
It was not omitted, but executed.

The important thing for you is now to understand that in your own working, the factor (t+C) is not correct (too much).

Perhaps it helps if you look at ##\displaystyle \int_{t_0}^{t_1} 2 \,dt = 2 t |_{t_0}^{t_1} = 2( t_1 - t_0 )##
In other words: the 2t in [(3t2)/2 + 2t] is a primitive of 2 dt
(and the 3t2)/2 is the primitive of 3t)

prakhargupta3301 said:
't+C' is the integral of dt)
There is no such thing as an integral of "dt".
∫dt = ∫1.dt is the integral of 1 with respect to t.
Likewise, ∫(3t+2).dt is the integral of 3t+2 with respect to t.

Secondly, you only need a constant of integration in indefinite integrals. It represents the fact that you do not know both bounds. Where you do know the bounds, the constant would be the same at each end of the range, and so cancels out:
ab2x.dx=[x2+C]ab=(b2+C)-(a2+C).

∫ dt = t + C or ∫ dt = Δt

Thank you BvU, Haruspex and Dr Dr news, because if you my concept is clear.

## 1. What are integrals and why are they important in science?

Integrals are mathematical tools used to find the area under a curve or the accumulation of a quantity over a given interval. They are important in science because they allow us to solve problems involving continuous variables, such as velocity, acceleration, and volume.

## 2. What are the common steps to solve an integral?

The common steps to solve an integral are: 1) identify the type of integral (definite or indefinite), 2) use appropriate integration rules (such as power rule, substitution, or integration by parts), 3) evaluate the integral, and 4) add any necessary constants or limits.

## 3. What are some common mistakes to avoid while solving integrals?

Some common mistakes to avoid while solving integrals include: forgetting to add the constant of integration, using incorrect integration rules, making algebraic errors, or not simplifying the final answer.

## 4. How can I improve my skills in solving integrals?

To improve your skills in solving integrals, practice regularly and review different integration techniques. It can also be helpful to understand the concept of antiderivatives and how they relate to integrals. Additionally, make sure to check your work and identify any mistakes to learn from them.

## 5. Are there any tips or tricks for solving integrals more efficiently?

Yes, there are several tips and tricks for solving integrals more efficiently, such as using symmetry to simplify the integral, looking for substitutions that can make the integral easier, and using integration tables or software programs. It is also helpful to break down complex integrals into smaller, more manageable parts.