Kyle asked in Science & MathematicsMathematics · 5 years ago

# u sub problem with definite integral.?

I solved the above problem to be integral 3u^(1/3) from 1 to 8 and put u as the 8y-5y^2+4y^3+1 but when I put the limits into this I get

3(1793^(1/3))-6

MML says it should be 3 but the only way I can figure out how they got 3 is by putting the limits 8 and 1 into 3u^(1/3). Is that how you solve this type of problem because I thought after finding u you were supposed to evaluate as F(b)-F(a).

Relevance
• 5 years ago

when substituting you have to substitute the integration limits, too

if u = 8y-5y^2+4y^3+1, then

when y = 0, u = 1

when y = 1, u = 8 - 5 + 4 + 1 = 8

thus if the primitive is 3u^(1/3), the integral is

3*8^(1/3) - 3*1^(1/3) = 6 - 3 = 3

hope this helps