# Fundamental Theorem of Calculus

• March 9th 2009, 03:35 PM
Macleef
Fundamental Theorem of Calculus
All my answers for the following questions are wrong, so please point out where I made the mistakes and how I could fix them or correct them for me! Thank you.

Evaluate the given definite integral using the fundamental theorem of calculus.

1) $\int_{-1}^1 (2u^{\frac {1}{3}} - u^{\frac {2}{3}} \, du$

$= \int_{-1}^1 \frac{2}{4} u^{\frac{1}{4}} - \frac{3}{5} u^{\frac{5}{3}} |_{-1}^1$

$= - \frac{1}{10} - \frac{1}{10}$

$= - \frac {1}{5}$

2) $\int_{1}^2 (2x - 4)^4 \, dx$

$= \frac{1}{5} (2x - 4)^5 \, |_1^2$

$= \frac{1}{5} (4 - 4)^5 - \frac{1}{5} (2 - 4)^5$

$= 6.4$

3) $\int_0^4 \frac{1}{\sqrt{6t +1}} \, dt$

$= \int_0^4 (6t + 1)^{- \frac{1}{2}} \, dt$

$= 2(6t + 1)^{\frac{1}{2}} |_0^4$

$= 8$

4) $\int_0^1 (x^3 + x) \sqrt{x^4 + 2x^2 + 1} \, dx$

$u = x^4 + 2x^2 + 1$

$du = 4x^3 + 4x \, dx$

$\int_0^1 (x^3 + x)u \frac{1}{4(x^3 + x)} \, du$

$= \frac{1}{8}u|_1^4$

$= \frac{15}{8}$

5) $\int_2^{e + 1} \frac{x}{x - 1} \, dx$

$u = x$

$du = dx$

$dv = (x - 1)^{-1} \, dx$

$v = x - 1$

$= x(x - 1)|_2^{e+1} + \int_2^{e + 1} x \, dx$

$= x^2 - x + \frac{1}{2}x^2 |_2^{e + 1}$

$= \frac{3}{2}(e + 1)^2 - (e + 1) + 3$

1) $-\frac{6}{5}$

2) $3.2$

3) $\frac{4}{3}$

4) $\frac{7}{6}$

5) e

P.S. How do you make that line "|" longer?
• March 10th 2009, 01:57 AM
Integration
Hello Macleef

Thanks for showing us your working - this must have taken you quite a long time!

I've corrected the first mistakes you've made in each case - I hope you can finish them now.

Quote:

Originally Posted by Macleef
All my answers for the following questions are wrong, so please point out where I made the mistakes and how I could fix them or correct them for me! Thank you.

Evaluate the given definite integral using the fundamental theorem of calculus.

1) $\int_{-1}^1 (2u^{\frac {1}{3}} - u^{\frac {2}{3}} \, du$

$= \int_{-1}^1 \frac{2}{4} \color{red}u^{\frac{1}{4}}\color{black} - \frac{3}{5} u^{\frac{5}{3}} |_{-1}^1$

$\color{red}\int u^{\frac{1}{3}}du = \frac{3}{4}u^{\frac{4}{3}}$

...

2) $\int_{1}^2 (\color{red}2\color{black}x - 4)^4 \, dx$

$= \frac{1}{5}\color{red}\times\frac{1}{2}\color{blac k} (2x - 4)^5 \, |_1^2$

...

3) $\int_0^4 \frac{1}{\sqrt{6t +1}} \, dt$

$= \int_0^4 (\color{red}6\color{black}t + 1)^{- \frac{1}{2}} \, dt$

$= 2\color{red}\times\frac{1}{6}\color{black}(6t + 1)^{\frac{1}{2}} |_0^4$

...

4) $\int_0^1 (x^3 + x) \sqrt{x^4 + 2x^2 + 1} \, dx$

$u = x^4 + 2x^2 + 1$

$du = 4x^3 + 4x \, dx$

$\int_0^1 (x^3 + x)\color{red}\sqrt{u}\color{black} \frac{1}{4(x^3 + x)} \, du$

...

5) $\int_2^{e + 1} \frac{x}{x - 1} \, dx$

$\color{red}= \int_2^{e+1}\left(1+\frac{1}{x-1}\right)dx\color{black}$

$\color{red}= \Big[x + \text{ln}|x-1|\Big]_2^{e+1}$

...

P.S. How do you make that line "|" longer?

$\color{red}|,\big|, \Big|, \bigg|, \Bigg|$