Comprehensive Guide To Solving Integrals And Finding Area Under Curves
Hey everyone! Today, we're diving deep into the fascinating world of calculus, specifically tackling some tricky integral problems and finding the area under a curve. Whether you're a student grappling with calculus concepts or just someone who loves the beauty of mathematics, this guide is for you. We'll break down each problem step-by-step, making sure you understand the why behind the how. Let's get started!
1. Evaluating the Integral of ∫x³e²ˣ dx
Let's kick things off with a classic integral that often appears in calculus courses: evaluating the integral of ∫x³e²ˣ dx. This integral requires a technique called integration by parts, which is a powerful tool for handling integrals involving products of functions. Now, you might be thinking, "Integration by parts? Sounds complicated!" But don't worry, we'll break it down into manageable steps.
Understanding Integration by Parts
Before we jump into the specifics of our problem, let's quickly recap the integration by parts formula. It states:
∫u dv = uv - ∫v du
Where:
- u is a function we choose to differentiate.
- dv is the remaining part of the integral that we choose to integrate.
- du is the derivative of u.
- v is the integral of dv.
The key to successfully using integration by parts lies in choosing the right u and dv. A helpful mnemonic for this is LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. This acronym gives you a general order of preference for choosing u. In our case, we have an algebraic function (x³) and an exponential function (e²ˣ). According to LIATE, we should choose the algebraic function as u.
Step-by-Step Solution for ∫x³e²ˣ dx
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Choose u and dv:
Let u = x³ and dv = e²ˣ dx
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Find du and v:
Differentiate u: du = 3x² dx
Integrate dv: v = ∫e²ˣ dx = (1/2)e²ˣ
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Apply the Integration by Parts Formula:
∫x³e²ˣ dx = x³(1/2)e²ˣ - ∫(1/2)e²ˣ(3x²) dx
Simplify: ∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/2)∫x²e²ˣ dx
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Repeat Integration by Parts:
Notice that we still have an integral involving a product of functions (x²e²ˣ). This means we need to apply integration by parts again. Let's choose:
u = x² and dv = e²ˣ dx
Then:
du = 2x dx and v = (1/2)e²ˣ
Applying the formula:
∫x²e²ˣ dx = x²(1/2)e²ˣ - ∫(1/2)e²ˣ(2x) dx
Simplify: ∫x²e²ˣ dx = (1/2)x²e²ˣ - ∫xe²ˣ dx
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Repeat Integration by Parts (Again!):
Yes, we need to do it one more time! Let:
u = x and dv = e²ˣ dx
Then:
du = dx and v = (1/2)e²ˣ
Applying the formula:
∫xe²ˣ dx = x(1/2)e²ˣ - ∫(1/2)e²ˣ dx
Simplify: ∫xe²ˣ dx = (1/2)xe²ˣ - (1/4)e²ˣ
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Substitute Back and Simplify:
Now, we need to substitute all the results back into our original equation:
∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/2)[(1/2)x²e²ˣ - ∫xe²ˣ dx]
∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/2)∫xe²ˣ dx
∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/2)[(1/2)xe²ˣ - (1/4)e²ˣ]
∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/4)xe²ˣ - (3/8)e²ˣ + C
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Final Answer:
The integral of ∫x³e²ˣ dx is (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/4)xe²ˣ - (3/8)e²ˣ + C, where C is the constant of integration.
Key Takeaway: Integration by parts can be a bit tedious, especially when you need to apply it multiple times. But by carefully choosing u and dv and systematically working through the steps, you can conquer even the most challenging integrals!
2. Evaluating the Definite Integral: ∫₀¹ 2x/(x²+1)² dx
Alright, let's move on to our next challenge: evaluating the definite integral ∫₀¹ 2x/(x²+1)² dx. This integral looks a bit different from the previous one, and we'll tackle it using a technique called u-substitution. U-substitution is your best friend when you spot a function and its derivative (or a multiple of its derivative) within the integral. Let's see how it works!
Understanding U-Substitution
The core idea behind u-substitution is to simplify the integral by replacing a part of the integrand with a new variable, u. This often transforms the integral into a more manageable form. The key steps are:
- Choose a suitable expression to be u.
- Find the derivative of u, which is du.
- Rewrite the integral in terms of u and du.
- Evaluate the integral with respect to u.
- Substitute back the original expression for u.
Step-by-Step Solution for ∫₀¹ 2x/(x²+1)² dx
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Choose u:
Look at the integral. We see (x²+1) in the denominator, and its derivative is 2x, which also appears in the integral. This is a clear sign that u-substitution will work. Let's choose:
u = x² + 1
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Find du:
Differentiate u with respect to x:
du = 2x dx
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Change the Limits of Integration:
Since we're dealing with a definite integral, we need to change the limits of integration to match our new variable u. When x = 0, u = 0² + 1 = 1. When x = 1, u = 1² + 1 = 2. So, our new limits of integration are 1 and 2.
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Rewrite the Integral:
Substitute u and du into the integral:
∫₀¹ 2x/(x²+1)² dx = ∫₁² 1/u² du = ∫₁² u⁻² du
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Evaluate the Integral:
Now we have a much simpler integral to evaluate:
∫₁² u⁻² du = [-u⁻¹]₁² = [-1/u]₁²
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Apply the Limits of Integration:
Substitute the limits of integration:
[-1/u]₁² = (-1/2) - (-1/1) = -1/2 + 1 = 1/2
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Final Answer:
The definite integral ∫₀¹ 2x/(x²+1)² dx equals 1/2.
Key Takeaway: U-substitution is a powerful technique for simplifying integrals when you can identify a function and its derivative within the integrand. Don't forget to change the limits of integration when dealing with definite integrals!
3. Finding the Area Under the Curve y = h(x) from x = 1 to x = 4
Last but not least, let's tackle the problem of finding the area under the curve y = h(x) from x = 1 to x = 4. This is a classic application of definite integrals. The area under a curve between two points is given by the definite integral of the function between those points. So, we're essentially looking for ∫₁⁴ h(x) dx.
The Fundamental Theorem of Calculus
To find the area, we'll rely on the Fundamental Theorem of Calculus. This theorem connects differentiation and integration and tells us that if H(x) is an antiderivative of h(x) (meaning H'(x) = h(x)), then:
∫ₐᵇ h(x) dx = H(b) - H(a)
In simpler terms, the definite integral of a function from a to b is equal to the difference between the antiderivative of the function evaluated at b and the antiderivative evaluated at a.
Applying the Theorem
The problem states that we need to find the area under the curve y = h(x) from x = 1 to x = 4, but it doesn't give us the explicit function h(x). Instead, it likely provides us with the antiderivative, H(x), or enough information to determine it.
Let's assume, for the sake of illustration, that we are given the antiderivative H(x) = x⁴ - 2x³ + x² + 5. (In a real problem, you would either be given H(x) directly or have to find it by integrating h(x)).
Step-by-Step Solution
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Identify the Antiderivative H(x):
We've assumed H(x) = x⁴ - 2x³ + x² + 5.
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Evaluate H(x) at the Limits of Integration:
H(4) = (4)⁴ - 2(4)³ + (4)² + 5 = 256 - 128 + 16 + 5 = 149
H(1) = (1)⁴ - 2(1)³ + (1)² + 5 = 1 - 2 + 1 + 5 = 5
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Apply the Fundamental Theorem of Calculus:
∫₁⁴ h(x) dx = H(4) - H(1) = 149 - 5 = 144
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Final Answer:
The area under the curve y = h(x) from x = 1 to x = 4 is 144 square units.
Key Takeaway: The Fundamental Theorem of Calculus is the cornerstone of evaluating definite integrals and finding areas under curves. Knowing the antiderivative of the function is crucial!
Conclusion: Mastering the Art of Integration
Alright, guys, we've covered a lot of ground in this comprehensive guide! We tackled a tricky integral using integration by parts (∫x³e²ˣ dx), conquered a definite integral with u-substitution (∫₀¹ 2x/(x²+1)² dx), and found the area under a curve using the Fundamental Theorem of Calculus. These are fundamental techniques in calculus, and mastering them will set you up for success in your mathematical journey.
Remember, practice makes perfect! The more you work through these types of problems, the more comfortable and confident you'll become. So, keep those pencils moving, and don't be afraid to ask for help when you need it. Happy integrating!
