Question

I'm stuck on the following Definite Integral problem, and would really appreciate someone taking a look at my work and walking me through understanding my error(s).

*I'll post the problem as the first comment.

Any and all help is greatly appreciated!

Answer 1

where is the problem?

Answer 2

The problem:
\[\int\limits_{0}^{1}\sqrt[3]{(1+7x})dx\]

Answer 3

You guys are too fast lol! :) I've literally had the question open for less than 45 seconds.

Answer 4

thanks

Answer 5

What did you get as your answer @amonoconnor ?

Answer 6

I got 3//28ths, but my books says that answer is 45/28. :/

Answer 7

Yeap, the answer is 45/28. Did you do this with substitution?

Answer 8

I just reached the same thing 45/28

Answer 9

This is how I did, using substitution.\[\int\limits_{0}^{1}\sqrt[3]{7x+1}dx\] Substitute \[u=7x+1-> du=7dx\]
This gives a new lower bound: \[u=1+7*0=1\] and upper bound \[u=1+7*1=8\] So now you have \[\frac{ 1 }{ 7 }\int\limits_{1}^{8}\sqrt[3]{u}du \] Here we want to apply the fundamental theorem of calculus. The antiderivative of \[\sqrt[3]{u}=\frac{ 3u^{4/3} }{ 4 }\]So you get: \[\frac{ 1 }{ 7 }\left[ \frac{ 3u^{4/3} }{ 4 } \right]_1^8=\frac{ 45 }{28}\]

Answer 10

Here's what I did:
\[u = (1+7x)\]
\[du = 7dx\]
\[dx = \frac{ 1 }{ 7 }du\]
\[\int\limits_{0}^{1}\sqrt[3]{(1+7x)}dx = \int\limits_{0}^{1}(u^{1/3})*\frac{1}{7}du\]
\[= \frac{ 1 }{ 7 }[\frac{3}{4}u^{4/3}], 0/1\]
\[= [\frac{3}{28}\sqrt[3]{(1+7x)^4}], 0/1\]
\[= [\frac{3}{28}\sqrt[3]{28}\]

Answer 11

Yeap, you need to change the bounds, when you make a substitution, thats the mistake.

Answer 12

Gotta... Ugg! I feel so dumb, that makes a ton of sense! Thank you!

Answer 13

You're welcome :)

Answer 14

sorry i couldn't be more help

Answer 15

No worries! We all do we can on OS, and sometimes hard stuff floats around, or we just haven't learned yet... I was there a lot in my past, and still have plenty of those moments:)