You forgot to change the limits.karush said:W.8.1.18
$$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx $$
$u=x^4-2x$ $du=4x^3-2 dx =2(2 x^3-1) dx$
$$\frac{1}{2}\int_{-1}^{1} u^6\,du $$
Is this ok
No, what you have done is correct. However, the limits should have changed. You're integrating the first integral between $x=-1$ and $x=1$. When you make the sub $u = x^4-2x $ your limits change! So $u = x^4-2x$ and when $x = -1$ you have $u = (-1)^4-2(-1) = 1+2 = 3$ and when $x = 1$ you have $u = (1)^4-2(1) = 1-2 = -1$. So you should have got:$$\displaystyle \frac{1}{2} \int_3^{-1} u^6 \,du$$karush said:not sure what you mean, should $u$ be something else?
Integral substitution, also known as u-substitution, is a method used to simplify and solve integrals that involve a function within another function. It involves replacing the inner function with a new variable, u, to make the integral easier to solve.
Integral substitution is useful because it allows us to solve complex integrals that would otherwise be difficult or impossible to solve using other methods. It also helps to make the integration process more efficient and organized.
You can use integral substitution when you have an integral that involves a function within another function, and the derivative of the inner function is also present in the integral. This method is particularly useful when the integral involves trigonometric functions, exponential functions, and logarithmic functions.
The steps for using integral substitution are as follows:
Some common mistakes to avoid when using integral substitution include forgetting to substitute the derivative of u back into the integral, using the wrong substitution for u, and not properly adjusting the limits of integration when substituting back in the original variable. It's important to carefully follow the steps and double check your work to avoid these errors.