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Consider the function $ f(x) = (2x-1)^6$. I want to find the integral. Intuitively to me, it would be $F(x) = \dfrac{1}{7} (2x-1)^7 + c$, but for some reason you have to add $\dfrac{1}{a} = \dfrac{1}{2}$ too. I have never really understood why. Why?

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Hint: $2x+1$ is not $x$ –  UrošSlovenija Apr 4 '13 at 17:29
    
You should differentiate your result and you'll see why you have to divide your result by 2. –  noobProgrammer Apr 4 '13 at 17:31

4 Answers 4

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Correction: you need to "multiply" the integral by $\dfrac 12$.

Try differentiating $\quad \dfrac 17(2x- 1)^7 + C$:

$$\dfrac d{dx}\left(\frac 17(2x - 1\right)^7 + C = (2x - 7)^6 \cdot \frac d{dx}\left(2x - 7\right) = 2(2x - 1)^6$$

You don't end up with the original integrand, because we need to use the chain rule to differentiate.

To this end, you can think of integrating $(2x - 1)^6$ by letting $\color{red}{\bf u = 2x - 1}$, $du = 2\,dx \implies \color{blue}{\bf dx = \frac 12 \,du}$. Then

$$\int \color{red}{\bf (2x - 1)}{\bf ^6} \,\color{blue}{\bf dx} = \int (\color{red}{\bf u}^{\bf 6}) \color{blue}{\bf \frac 12 du} = {\bf \dfrac{1}{2} \int u^6\,du}$$

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Nice Amy ;-)))) –  Babak S. Apr 4 '13 at 17:40
    
Thanks, Babak! ;-) –  amWhy Apr 4 '13 at 17:41

Think of inverting this via the Chain rule. Note that $$ \frac{dF}{dx} = \frac{d}{dx} \frac{1}{7} (2x-1)^7 = \frac{1}{7} 7 (2x-1)^6 \frac{d(2x-1)}{dx} = \frac{1}{7} 7 (2x-1)^6 \cdot 2. $$

That last $2$ needs to be counterbalanced in integration, that's where the factor of $1/2$ comes from.

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Differentiate your answer. You will see that an extra 2 appears in the denominator as per chain rule. So you need to divide by two so as to get the same function as the derivative.

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The answer concerns the chain rule. Notice that $F'(x) = (2x-1)^6 \cdot \frac{d}{dx} (2x-1)$.

In general, you cannot integrate $g(x)^6$ by treating $g(x)$ like $x$. You can, however, integrate $\int g(x)^6 g'(x) dx = \frac{1}{7} g(x)^7$. This is the basis of the technique of $u$-substitution.

Applying $u$-substitution to this problem, we would set $u = 2x-1$. Then, $du = 2dx$.

$$\int (2x-1)^6 dx = \frac{1}{2} \int u^6 du = \frac{1}{14} u^7 + C = \frac{1}{14} (2x-1)^7 + C$$

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