Math  /  Calculus

Questionoblem \#15: Consider the following integrals. (i) 062x1/4dx\int_{0}^{6} \frac{2}{x^{1 / 4}} d x (ii) 02x4dx\int_{0}^{\infty} \frac{2}{x^{4}} d x (iii) 04xdx\int_{0}^{\infty} 4^{x} d x (iv) 64xdx\int_{-\infty}^{6} 4^{x} d x
Determine if the above integrals Converge (1) or Diverge (2). So, for example, if you think that the answers, in the above order, are Converge,Diverge,Diverge,Converge then you would enter ' 1,2,2,11,2,2,1 ' into the answer box below (without the quotes).

Studdy Solution

STEP 1

What is this asking? We need to figure out if each of these integrals converges to a specific value, or diverges (meaning it goes to infinity or doesn't settle on a single value). Watch out! Remember that infinity isn't a number we can just plug in!
We need to take limits.
Also, be super careful around 00, as some functions can get really weird there.

STEP 2

1. Analyze the first integral
2. Analyze the second integral
3. Analyze the third integral
4. Analyze the fourth integral

STEP 3

Let's rewrite the integrand 2x1/4\frac{2}{x^{1/4}} as 2x1/42x^{-1/4} to make it easier to use the power rule for integration.
This makes our integral 062x1/4dx\int_{0}^{6} 2x^{-1/4} dx.

STEP 4

The power rule says xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C.
Here, our nn is 1/4-1/4, so n+1n+1 is 1/4+1=3/4-1/4 + 1 = 3/4.
So, the integral becomes 2x3/43/42 \cdot \frac{x^{3/4}}{3/4}.
Remember to keep the 22 out front!

STEP 5

Dividing by 3/43/4 is the same as multiplying by 4/34/3, so our integral evaluates to 243x3/4=83x3/42 \cdot \frac{4}{3}x^{3/4} = \frac{8}{3}x^{3/4}.

STEP 6

We need to evaluate this from 00 to 66.
That means we plug in 66 and 00 and subtract the results: 83(63/4)83(03/4)\frac{8}{3}(6^{3/4}) - \frac{8}{3}(0^{3/4}).
Since 03/40^{3/4} is just 00, the second term disappears!
We're left with 83(63/4)\frac{8}{3}(6^{3/4}), which is a perfectly normal, finite number.

STEP 7

This integral **converges**.

STEP 8

We can rewrite 2x4\frac{2}{x^4} as 2x42x^{-4}.
This makes our integral 02x4dx\int_{0}^{\infty} 2x^{-4} dx.
Uh oh, infinity!

STEP 9

Since we're integrating to infinity, we need to take a limit: limb0b2x4dx\lim_{b\to\infty} \int_{0}^{b} 2x^{-4} dx.

STEP 10

Using the power rule, the integral becomes 2x332 \cdot \frac{x^{-3}}{-3}, which simplifies to 23x3-\frac{2}{3x^3}.

STEP 11

We want limb(23b323(0)3)\lim_{b\to\infty} \left( -\frac{2}{3b^3} - -\frac{2}{3(0)^3} \right).
Hold on!
We can't divide by zero.
This integral is improper at *both* 00 and infinity.
We need to break it up!
Let's pick 11 as our breaking point: lima0+a12x4dx+limb1b2x4dx\lim_{a\to 0^+} \int_a^1 2x^{-4}dx + \lim_{b\to\infty} \int_1^b 2x^{-4}dx.

STEP 12

The first part becomes lima0+(23(1)323a3)\lim_{a\to 0^+} \left( -\frac{2}{3(1)^3} - -\frac{2}{3a^3} \right).
As aa gets super close to 00, 23a3\frac{2}{3a^3} becomes super huge, so this limit goes to infinity.

STEP 13

Since the first part diverges, the whole integral **diverges**.
We don't even need to check the second part!

STEP 14

We have 04xdx\int_{0}^{\infty} 4^x dx, which means we need a limit: limb0b4xdx\lim_{b\to\infty} \int_{0}^{b} 4^x dx.

STEP 15

Remember that axdx=axln(a)\int a^x dx = \frac{a^x}{\ln(a)}.
So, 4xdx=4xln(4)\int 4^x dx = \frac{4^x}{\ln(4)}.

STEP 16

We get limb(4bln(4)40ln(4))\lim_{b\to\infty} \left( \frac{4^b}{\ln(4)} - \frac{4^0}{\ln(4)} \right).
As bb goes to infinity, so does 4b4^b.

STEP 17

This integral **diverges**.

STEP 18

We have 64xdx\int_{-\infty}^{6} 4^x dx, so we need a limit: limaa64xdx\lim_{a\to -\infty} \int_{a}^{6} 4^x dx.

STEP 19

The integral is 4xln(4)\frac{4^x}{\ln(4)}.

STEP 20

We have lima(46ln(4)4aln(4))\lim_{a\to -\infty} \left( \frac{4^6}{\ln(4)} - \frac{4^a}{\ln(4)} \right).
As aa goes to negative infinity, 4a4^a goes to 00.
So, we're left with 46ln(4)\frac{4^6}{\ln(4)}, a finite number.

STEP 21

This integral **converges**.

STEP 22

Our final answer is **1,2,2,1**.

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