Definite Integral Example

[Jason76]

\(\displaystyle \int_{1}^{8} x \sqrt{x + 1} dx\)

\(\displaystyle \int_{1}^{8} x (x + 1)^{1/2} dx\)

\(\displaystyle \int_{1}^{8} x (u)^{1/2} dx\)


\(\displaystyle u = x + 1\)

\(\displaystyle du = dx\) - This cannot be manipulated using fraction multipication.

So

\(\displaystyle u = x + 1\)

\(\displaystyle x = u - 1\) solve for x

Replace whole equation based on new info:

\(\displaystyle \int_{1}^{8} (u - 1) (u)^{1/2}\)

\(\displaystyle \int_{1}^{8} u^{3/2} - u^{1/2}\) - using the distributive property

Problem here: There should be new limits of integration. How do I calculate them?

 

[stapel]

\(\displaystyle \displaystyle{\int_{1}^{8}\, x\, \sqrt{x\, +\, 1}\, dx}\)

\(\displaystyle u\, =\, x\, +\, 1\)

....There should be new limits of integration. How do I calculate them?

When x = 1, what is u?
When x = 8, what is u?

Further info and examples available here.

 

[Jason76]

\(\displaystyle \int_{1}^{8} x \sqrt{x + 1} dx\)

\(\displaystyle \int_{1}^{8} x (x + 1)^{1/2} dx\)

\(\displaystyle \int_{1}^{8} x (u)^{1/2} dx\)


\(\displaystyle u = x + 1\)

\(\displaystyle du = dx\) - This cannot be manipulated using fraction multipication.

So

Make new limits:

\(\displaystyle u(8) = (8) + 1 = 9\)

\(\displaystyle u(1) = (1) + 1 = 2\)

\(\displaystyle u = x + 1\) - solve this for x

\(\displaystyle x = u - 1\)

Replace whole equation based on new info:

\(\displaystyle \int_{2}^{9} (u - 1) (u)^{1/2}\)

\(\displaystyle \int_{2}^{9} u^{3/2} - u^{1/2}\) - using the distributive property

\(\displaystyle \rightarrow \dfrac{u^{5/2}}{\dfrac{5}{2}} - \dfrac{u^{3/2}}{\dfrac{3}{2}}\)

 

[pka]

\(\displaystyle \int_{1}^{8} x \sqrt{x + 1} dx\)

\(\displaystyle \int_{1}^{8} x (x + 1)^{1/2} dx\)

\(\displaystyle \int_{1}^{8} x (u)^{1/2} dx\)


\(\displaystyle u = x + 1\)

\(\displaystyle du = dx\) - This cannot be manipulated using fraction multipication.

So

Make new limits:

\(\displaystyle u(8) = (8) + 1 = 9\)

\(\displaystyle u(1) = (1) + 1 = 2\)

\(\displaystyle u = x + 1\) - solve this for x

\(\displaystyle x = u - 1\)

Replace whole equation based on new info:

\(\displaystyle \int_{2}^{9} (u - 1) (u)^{1/2}\)

\(\displaystyle \int_{2}^{9} u^{3/2} - u^{1/2}\) - using the distributive property

Problem here: There should be new limits of integration. How do I calculate them?

The short answer is no.
Just use \(\displaystyle \int_{2}^{9} u^{3/2} - u^{1/2}\).

 

[Jason76]

Solving this without making new limits (also optional)

\(\displaystyle \int_{1}^{8} x \sqrt{x + 1} dx\)

\(\displaystyle x \sqrt{x + 1} dx\)

\(\displaystyle \int x (x + 1)^{1/2} dx\)

\(\displaystyle \int x (u)^{1/2} dx\)


\(\displaystyle u = x + 1\)

\(\displaystyle du = dx\) - This cannot be manipulated using fraction multipication.

So

\(\displaystyle u = x + 1\) - solve this for x

\(\displaystyle x = u - 1\)

Replace whole equation based on new info:

\(\displaystyle \int (u - 1) (u)^{1/2}\)

\(\displaystyle \int u^{3/2} - u^{1/2}\) - using the distributive property

\(\displaystyle \rightarrow \dfrac{u^{5/2}}{\dfrac{5}{2}} - \dfrac{u^{3/2}}{\dfrac{3}{2}}\)

\(\displaystyle \rightarrow \dfrac{2}{5} u^{5/2} - \dfrac{2}{3} u^{3/2}\)

\(\displaystyle \rightarrow \dfrac{2}{5} (x + 1)^{5/2} - \dfrac{2}{3} (x + 1)^{3/2} + C\)

\(\displaystyle [\dfrac{2}{5} ((8) + 1)^{5/2} - \dfrac{2}{3} ((8) + 1)^{3/2}] - [\dfrac{2}{5} ((1) + 1)^{5/2} - \dfrac{2}{3} ((1) + 1)^{3/2}]\)

\(\displaystyle [\dfrac{486}{5} - 18] - [?]\)

 

[lookagain]

\(\displaystyle \int_{1}^{8} x \sqrt{x + 1} dx\)

\(\displaystyle \int_{1}^{8} x (x + 1)^{1/2} dx\)

\(\displaystyle \int_{1}^{8} x (u)^{1/2} dx\)


\(\displaystyle u = x + 1\)

\(\displaystyle du = dx\) - This cannot be manipulated using fraction multipication.

So

Make new limits:

\(\displaystyle u(8) = (8) + 1 = 9\)

\(\displaystyle u(1) = (1) + 1 = 2\)

\(\displaystyle u = x + 1\) - solve this for x

\(\displaystyle x = u - 1\)

Replace whole equation based on new info:


I (lookagain) will add some missing characters/symbols in here:


\(\displaystyle \int_{2}^{9} (u - 1) (u)^{1/2}\)du

\(\displaystyle \int_{2}^{9} (u^{3/2} - u^{1/2})\) du \(\displaystyle \ \ \ \ \) - using the distributive property

\(\displaystyle \rightarrow \dfrac{u^{5/2}}{\frac{5}{2}} - \dfrac{u^{3/2}}{\frac{3}{2}}\bigg|_{2}^{9}\)

\(\displaystyle \dfrac{2}{5}u^{5/2} \ - \ \dfrac{2}{3}u^{3/2} \ \bigg|_{2}^{9}\)

.

 

[Jason76]

\(\displaystyle u = x + 1\) - solve this for x

\(\displaystyle x = u - 1\)

Replace whole equation based on new info:

\(\displaystyle \int (u - 1) (u)^{1/2}\)

\(\displaystyle \int u^{3/2} - u^{1/2}\) - using the distributive property

The above step is needed regardless of whether limits of integration are changed, assuming fraction multiplication won't work, or the dx matches what is in the original problem.

 

[stapel]

The above step is needed regardless of whether limits of integration are changed, assuming fraction multiplication won't work, or the dx matches what is in the original problem.

What do you mean by "assuming fraction multiplication won't work"? Why would multiplication or fractions not function somehow? Is the above statement meant to be a question of some sort? What do you want?

Seriously, we really can't read your mind.