MHB What is the integral of 2^(2x)? tonight, exam is tomorrow

  • Thread starter Thread starter mathnoob12345
  • Start date Start date
  • Tags Tags
    Exam Integral
Click For Summary
The integral of 2^(2x) is calculated using the substitution method. The teacher confirms the answer as 2^(2x) / (2Ln(2)). By rewriting 2^(2x) as e^(2Ln(2)x), the integral can be simplified. The substitution u = 2Ln(2)x leads to a straightforward integration of e^u. This method effectively demonstrates the steps to arrive at the final result for the integral.
mathnoob12345
Messages
1
Reaction score
0
what is the integral of 2^(2x)? need help tonight, exam is tomorrow

teacher says the answer is: 2^(2x) / 2Ln(2)

why 2 times Ln(2)?
 
Physics news on Phys.org
Observe that:

$$2^{2x}=e^{\ln\left(2^{2x}\right)}=e^{2\ln\left(2\right)x}$$

Hence:

$$I=\int 2^{2x}\,dx=\int e^{2\ln\left(2\right)x}\,dx$$

Let:

$$u=2\ln\left(2\right)x\implies du=2\ln\left(2\right)\,dx$$

And so we have:

$$I=\frac{1}{2\ln\left(2\right)}\int e^u\,du$$

Can you proceed?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 16 ·
Replies
16
Views
874
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K