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Indefinite Integrals
In this tutorial, we will cover some elementary indefinite integral examples. Let's start with an example: Example 1: $$\int 3x^5 dx$$ The first step is to add one to the exponent. So, we add one to 5 and get 6. Next, we divide the coefficient (3) by the new exponent, 6. Simplifying, we get 1/2. Hence, the integral is: $$\frac{1}{2}x^6 + C$$ where C is the constant of integration. Always remember to add C at the end of integrating!
Indefinite Integrals 2
The first thing you want to do whenever you integrate a square root is to get rid of the square root and write it with a 1/2 exponent instead. For example: ∫√x dx = ∫x^(1/2) dx After you add 1 to the exponent you need to divide by the same so since we have an exponent of 3/2 we're gonna divide by 3/2. So: ∫x^(3/2) dx = 2/5 x^(5/2) + c And we can’t forget about our last step, you always need to add c:
Indefinite Integrals 3
First, we need to move the x from the denominator to the numerator: {$$\int{\\frac{1}{x^3}}dx = \\int{x^{-3}}dx = -\\frac{1}{2}x^{-2}$$} After integrating, we end up with a negative 3 exponent. The constant stays the same: {$$\\int{\\frac{1}{x^3}}dx = -\\frac{1}{2x^2} + C$$}.
U substitution
In this passage, we will cover some complex integral examples that require the use of u substitution. Let's consider the following integral: ∫ 6x(x² - 5)⁵ dx
To integrate this expression, we need to use substitution. Instead of writing the integral of 6x dx, we can replace it with 3du (since 6x dx = 3du). So, the Integral becomes: 3 ∫ u⁵ (x² - 5)⁵ du We can simplify this expression to: 3 ∫ u⁵ (u² - 14)⁵ du Since we have expressed the integral in terms of u, we can now integrate using power rule: ∫ u⁵ (u² - 14)⁵ du = ∫ u⁷ - 70u⁵ + 2450u³ - 34300u + 168070 du Finally, for every indefinite integral, we always add our constant 'C': ∫ 6x(x² - 5)⁵ dx = (3/8) (x² - 5)⁶ + C
U substitution 2
We are given: ∫ e^(x^2+5x) * (4x+10) dx We will use u-substitution to replace the exponent on top of the e with a u. Let u = x^2 + 5x Then du/dx = 2x + 5 And dx = (du / (2x + 5)) Substituting, we get: ∫ e^u * (4x+10) * (du / (2x + 5)) Factoring out (2x + 5) in the denominator, we get: ∫ e^u * (4x+10) * (du / (2x + 5)) = 2 ∫ e^u * (2x+5) * (du / (2x+5)) Cancelling out the (2x+5) terms, we get: = 2 ∫ e^u du Integrating, we get: = 2 e^u
Cosecant function: csc(x) = 1/sin(x) Natural logarithm function: ln(x) Now, let's move on to the example at hand: We have the integral of cos(x) - sin(x) with upper limit pi and lower limit 0. By plugging in pi wherever there is an x, we get: cos(pi) - sin(pi) - (cos(0) - sin(0)) Simplifying this further, we get: -1 - 0 - (1 - 0) Finally, our answer is:
Area under a curve introduction
Let's consider our lower point as 3 and our upper point as 7. Each rectangle touches the blue curve, which is -x2 + 10x - 21, and the bottom of every rectangle touches the green curve, which is x2 - 10x + 21 dx. The area under the curves is given by the integral of a, which is our lower point of intersection (3), to b, which is our top point of intersection (7), and to the function where all the tops of the rectangles touch.
Area under a curve
To find the area bounded by the curves, we will be using the following two equations: y = x2 - 6x + 9 y = x + 3 The first thing that needs to be done is to graph both equations to better understand which is on top and which is on the bottom, as well as where they intersect. The area bounded by the curves is equal to: 18 - (-17 / 6) = 125 / 6 Therefore, the exact area between the curve x + 3 and the curve x2 - 6x + 9 dx is equal to 125 / 6. This can be expressed as a definite integral with an upper limit of 6 and a lower limit of 1: ∫16 (x2 - 6x + 9) - (x + 3) dx
Integration by parts
The only possible method to solve this integral by hand is integration by parts. We have two parts to the integrals. We have f(x) and we have g'(x). The only thing we have left is ex , so the ex has to be our g'(x). We need to plug everything into the formula, so let's start with our f(x).
Our f(x) is equal to x , so we're going to plug in an x for f(x). Now let's go back to our example. We have x*ex , and this is being subtracted by the integral of ex. We use integration by parts to rewrite it in a much simpler way, which we can just integrate using basic integration. We can't forget to put our dx at the end of the integral, so notice now we have an integral that's much simpler.
Integration by parts - Hard example
The integral we are trying to solve is: ∫ exsin(x) dx This is a harder example than in the first video on integration by parts. We will find out in the end that indeed, we can solve this using integration by parts, so let's get started right away. We will go back to the formula: ∫u dv = uv - ∫v du The first step is to choose u and dv, such that after integrating the second term on the right-hand side, it will have a simpler form. Let's choose: u = sin(x), dv = exdx This means: du/dx = cos(x), v = ex Using the formula for integration by parts, we get: ∫ exsin(x) dx = -excos(x) + ∫ excos(x) dx We can simplify the second term using integration by parts again. Let's choose: u = cos(x), dv = exdx This means: du/dx = -sin(x), v = ex Using the formula for integration by parts, we get: ∫ excos(x) dx = exsin(x) - ∫ exsin(x) dx Substituting this back into our original equation, we get: ∫ exsin(x) dx = -excos(x) + exsin(x) - ∫ exsin(x) dx Adding ∫ exsin(x) dx to both sides, we get: 2∫ exsin(x) dx = -excos(x) + exsin(x) + C Therefore: ∫ exsin(x) dx = (-1/2)excos(x) + (1/2)exsin(x) + C To find the integral of the following expression: ∫ [ sin(x) * e^(ex) - cos(x) * e^(x) ] dx and on the right-hand side, we have: sin^x - cos^x We need to divide both sides by 2 + C. This example shows how integration by parts can be challenging.
I hope this helps you understand how to solve integrals using inverse trig functions.
Thank You So Much.
