Is the integral of 2x + x + 1 equal to (1)(3) + \frac{1}{2}(1)(3) + 1?

  • Thread starter Thread starter courtrigrad
  • Start date Start date
  • Tags Tags
    Integration
AI Thread Summary
The integral of 2x + x + 1 from 1 to 2 simplifies to 3x + 1, leading to the evaluation of the definite integral as 5.5. There is speculation that the original equation might have intended to be 2x^2 + x + 1, suggesting a focus on basic integral forms. Key integral formulas discussed include the integration of x and x^2 over a defined interval. The calculations confirm that the original equation's terms are correct, provided the first term is indeed 2x. The discussion emphasizes the importance of clarity in the expression of integrals.
courtrigrad
Messages
1,236
Reaction score
2
Is \int^2_1 2x + x + 1 dx= (1)(3) + \frac{1}{2}(1)(3) + 1?
 
Physics news on Phys.org
Why doncha tell us how u came up with those numbers...??

Daniel.
 
\int^2_1 2x + x + 1 dx = \int^2_1 3x + 1 dx = \left[\frac{3}{2}x^2 + x \right]^2_1
When you solve that i think u get 5.5

Though i suspect that original equation would be
\int^2_1 2x^2 + x + 1 dx or why would they give it to you in that form ... 2x + x ?
 
Last edited:
They were trying to illustrate the basic integrals

\int^b_a x dx = \frac{1}{2}(b-a)(b+a) and \int^b_a x^2 dx = \frac {1}{3}(b^3 - a^3)

\int^b_a c dx = c(b-a)

This is how I got those numbers
 
You answer is correct.
 
your answer is absolutely correct. (if you sure the first term is 2x instead of 2x^2 as ryoukomaru stated earlier)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top