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Tricky Substitutions Involving Radicals - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
When radicals or fractional powers are involved, substitute the expression being raised to the fractional power with w. Then, use the substitution method: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.
Let’s try a trickier problem. Here I have the integral of x plus 3 over 3 minus x to the 2/3. Now so far, you may have noticed that in the examples I have done, I have substituted for the thing inside the radical each time. Here there is no obvious radical but you have to remember that 3 minus x to the 2/3. That is a radical function.
A fractional power represents a radical. The fraction 1/3 represents the cube root, so this is something squared raised to the cube root of something squared. So I want to substitute for the 3 minus x the way I have been substituting for things inside the radical. So let me try that. W equals 3 minus x. And then I need a dw to replace my dx. So dw is –dx, and that means dx is –dw. I'm going to replace the dx with a -dw. I still have this x plus 3 to deal with. So I would have dealt with this and this but not this. So how do I get x plus 3?
It comes from this formula. I can solve for x I can add x to both sides and subtract w from both side. And I can get x equals 3 minus w. But that’s not x plus 3 x plus 3. X plus 3 I’m basically adding 3 to both sides of this. I would get 6 minus w. So I can replace the x plus 3 with 6 minus w. So let me do that so this x plus 3 becomes 6 minus w.
The dx becomes –dw, and the 3 minus x to the 2/3 becomes w to the 2/3. Just a little trick when you multiply a difference by a -1, you can flip the difference. Because -1 times 6 will be -6. -1 times –w would be +w. So this would be the same as w minus 6, I move the dw over a little bit. So this negative is gone because I just distributed over these two. And this w to the 2/3 I’m going to bring up to the top as w to the -2/3. So just a property of exponents I can bring that up. And then, let me distribute this over these two terms. And after I’ve done that, I’ll have just 2 powers of w that I can integrate really easily.
W times w to the negative 2/3 is w to the 1/3 minus 6 times w to the -2/3. So this is my new integral match easier because it just involves powers of w. First the integral of w to the 1/3. When I integrate this, the exponent goes up by 1 so from 1/3 to 4/3.
W to the 4/3, but then I divide by 4/3 but that’s the same as multiplying by ¾. Then I have a minus 6 and when I integrate w to the -2/3, I raise the exponent by 1. When you add 1 to the -2/3 you get 1./3 . So I get w to the 1/3. And I divide by 1/3. That’s the same as multiplying by 3.
Then I add c. So this is going to become ¾ w to the 4/3 minus 18w to the 1/3 plus c. All that remains to be done is to replace w with 3 minus x. So my final answer is going to be ¾, 3 minus x to the 4/3 minus 18 times 3 minus x to the 1/3 plus c. My final answer.
So remember when you see a radical function inside the integral, or a function with a fractional exponent, replace what’s inside the fraction exponent or the radical with w. That substitution may work for you.
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