Math Problem Statement

Evaluate the following (definite and indefinite) integrals: (a) ∫(e^x + 1/x) dx (b) ∫[1 to 2] (3u - 2)(u + 1) du (c) ∫ cos^3(θ) sin(θ) dθ (d) ∫[1 to 3] (1/z^2 + 1/(√z^3)) dz

Solution

I see you've uploaded an image with several integrals. I'll go ahead and evaluate each one.

(a) (ex+1x)dx\int \left(e^x + \frac{1}{x}\right) \, dx

To solve this, we can break it into two parts:

  1. exdx=ex\int e^x \, dx = e^x
  2. 1xdx=lnx\int \frac{1}{x} \, dx = \ln |x|

Thus, the solution is: ex+lnx+Ce^x + \ln |x| + C where CC is the constant of integration.

(b) 12((3u2)(u+1))du\int_1^2 \left( (3u - 2)(u + 1) \right) \, du

First, expand the integrand: (3u2)(u+1)=3u2+3u2u2=3u2+u2(3u - 2)(u + 1) = 3u^2 + 3u - 2u - 2 = 3u^2 + u - 2

Now, integrate term by term: 12(3u2+u2)du=[u3+u222u]12\int_1^2 (3u^2 + u - 2) \, du = \left[ u^3 + \frac{u^2}{2} - 2u \right]_1^2

Evaluating at the limits: =(23+2222(2))(13+1222(1))= \left( 2^3 + \frac{2^2}{2} - 2(2) \right) - \left( 1^3 + \frac{1^2}{2} - 2(1) \right) =(8+24)(1+122)= \left( 8 + 2 - 4 \right) - \left( 1 + \frac{1}{2} - 2 \right) =6(12)=6+12=132= 6 - \left( -\frac{1}{2} \right) = 6 + \frac{1}{2} = \frac{13}{2}

(c) cos3(θ)sin(θ)dθ\int \cos^3(\theta) \sin(\theta) \, d\theta

This is a standard trigonometric integral. We can use substitution: Let u=cos(θ)u = \cos(\theta), so that du=sin(θ)dθdu = -\sin(\theta) \, d\theta.

The integral becomes: u3(du)=u3du=u44+C=cos4(θ)4+C\int u^3 (-du) = -\int u^3 \, du = -\frac{u^4}{4} + C = -\frac{\cos^4(\theta)}{4} + C

(d) 13(1z2+1z3)dz\int_1^3 \left( \frac{1}{z^2} + \frac{1}{\sqrt{z^3}} \right) \, dz

We will evaluate each part:

  1. 1z2dz=1z\int \frac{1}{z^2} \, dz = -\frac{1}{z}
  2. 1z3dz=z32dz=2z\int \frac{1}{\sqrt{z^3}} \, dz = \int z^{-\frac{3}{2}} \, dz = \frac{2}{\sqrt{z}}

Now, evaluate from 1 to 3: 13(1z2+1z3)dz=[1z+2z]13\int_1^3 \left( \frac{1}{z^2} + \frac{1}{\sqrt{z^3}} \right) \, dz = \left[ -\frac{1}{z} + \frac{2}{\sqrt{z}} \right]_1^3

Evaluating at the limits: =(13+23)(1+2)= \left( -\frac{1}{3} + \frac{2}{\sqrt{3}} \right) - \left( -1 + 2 \right) =(13+23)1= \left( -\frac{1}{3} + \frac{2}{\sqrt{3}} \right) - 1 =13+231= -\frac{1}{3} + \frac{2}{\sqrt{3}} - 1 =43+23= -\frac{4}{3} + \frac{2}{\sqrt{3}}

Let me know if you need further clarification or more details on any of these!


Here are some related questions you might find interesting:

  1. How do you perform integration by parts?
  2. What methods can you use to evaluate improper integrals?
  3. How can you evaluate definite integrals with trigonometric identities?
  4. Can you explain how to solve integrals involving logarithmic and exponential functions?
  5. What are the common substitution techniques in integration?

Tip: Always check the domain of the functions in integrals, especially when they involve expressions like 1x\frac{1}{x} or z\sqrt{z}, to ensure proper handling of limits.

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Math Problem Analysis

Mathematical Concepts

Indefinite Integrals
Definite Integrals
Substitution in Integrals
Trigonometric Integration

Formulas

∫ e^x dx = e^x + C
∫ 1/x dx = ln|x| + C
∫ u^n du = (u^(n+1))/(n+1) + C for n ≠ -1
∫ cos^n(x)sin(x) dx = -cos^(n+1)(x)/(n+1) + C
∫ z^n dz = z^(n+1)/(n+1) + C for n ≠ -1

Theorems

Fundamental Theorem of Calculus
Power Rule for Integration
Logarithmic Rule for Integration

Suitable Grade Level

Undergraduate Calculus