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Math  /  Calculus

Question21. Find ∫1∞1xdx\int_{1}^{\infty} \frac{1}{x} \mathrm{dx}. a) 0 c) −∞-\infty b) 1 d) ∞\infty
22. Find ∫1∞2x3dx\int_{1}^{\infty} \frac{2}{x^{3}} d x. a) 0 c) 2 b) 1 d) ∞\infty
23. Find ∫0414−xdx\int_{0}^{4} \frac{1}{\sqrt{4-x}} d x. a) -4 b) 0 c) 4 d) does not exist.

Studdy Solution

STEP 1

1. We are dealing with improper integrals, which may converge to a finite value or diverge to infinity.
2. For each integral, we will determine convergence or divergence and compute the value if it converges.

STEP 2

1. Evaluate ∫1∞1x dx\int_{1}^{\infty} \frac{1}{x} \, dx.
2. Evaluate ∫1∞2x3 dx\int_{1}^{\infty} \frac{2}{x^3} \, dx.
3. Evaluate ∫0414−x dx\int_{0}^{4} \frac{1}{\sqrt{4-x}} \, dx.

STEP 3

To evaluate ∫1∞1x dx\int_{1}^{\infty} \frac{1}{x} \, dx, recognize it as an improper integral. Rewrite it with a limit:
∫1∞1x dx=lim⁡t→∞∫1t1x dx \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x} \, dx

STEP 4

Compute the integral ∫1t1x dx\int_{1}^{t} \frac{1}{x} \, dx:
∫1x dx=ln⁡∣x∣+C \int \frac{1}{x} \, dx = \ln|x| + C
Evaluate from 1 to t t :
[ln⁡∣x∣]1t=ln⁡∣t∣−ln⁡∣1∣=ln⁡t \left[ \ln|x| \right]_{1}^{t} = \ln|t| - \ln|1| = \ln t

STEP 5

Take the limit as t→∞ t \to \infty :
lim⁡t→∞ln⁡t=∞ \lim_{t \to \infty} \ln t = \infty
Since the limit is infinite, the integral diverges.
The answer for problem 21 is: ∞\boxed{\infty}

STEP 6

To evaluate ∫1∞2x3 dx\int_{1}^{\infty} \frac{2}{x^3} \, dx, rewrite it with a limit:
∫1∞2x3 dx=lim⁡t→∞∫1t2x3 dx \int_{1}^{\infty} \frac{2}{x^3} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{2}{x^3} \, dx

STEP 7

Compute the integral ∫1t2x3 dx\int_{1}^{t} \frac{2}{x^3} \, dx:
∫2x3 dx=−22x2+C=−1x2+C \int \frac{2}{x^3} \, dx = -\frac{2}{2x^2} + C = -\frac{1}{x^2} + C
Evaluate from 1 to t t :
[−1x2]1t=−1t2+112=1−1t2 \left[ -\frac{1}{x^2} \right]_{1}^{t} = -\frac{1}{t^2} + \frac{1}{1^2} = 1 - \frac{1}{t^2}

STEP 8

Take the limit as t→∞ t \to \infty :
lim⁡t→∞(1−1t2)=1 \lim_{t \to \infty} \left( 1 - \frac{1}{t^2} \right) = 1
Since the limit is finite, the integral converges.
The answer for problem 22 is: 1\boxed{1}

STEP 9

To evaluate ∫0414−x dx\int_{0}^{4} \frac{1}{\sqrt{4-x}} \, dx, note that the integrand becomes undefined at x=4 x = 4 . Rewrite it with a limit:
∫0414−x dx=lim⁡t→4−∫0t14−x dx \int_{0}^{4} \frac{1}{\sqrt{4-x}} \, dx = \lim_{t \to 4^-} \int_{0}^{t} \frac{1}{\sqrt{4-x}} \, dx

STEP 10

Compute the integral ∫0t14−x dx\int_{0}^{t} \frac{1}{\sqrt{4-x}} \, dx:
Using the substitution u=4−x u = 4 - x , du=−dx du = -dx , the integral becomes:
∫1u(−du)=−2u+C \int \frac{1}{\sqrt{u}} (-du) = -2\sqrt{u} + C
Evaluate from 0 to t t :
[−24−x]0t=−24−t+24−0=−24−t+4 \left[ -2\sqrt{4-x} \right]_{0}^{t} = -2\sqrt{4-t} + 2\sqrt{4-0} = -2\sqrt{4-t} + 4

STEP 11

Take the limit as t→4− t \to 4^- :
lim⁡t→4−(−24−t+4)=−2(0)+4=4 \lim_{t \to 4^-} \left( -2\sqrt{4-t} + 4 \right) = -2(0) + 4 = 4
Since the limit is finite, the integral converges.
The answer for problem 23 is: 4\boxed{4}

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