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Evaluate the following integrals. $$\int t^{2} e^{-t} d t$$

Short Answer

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Question: Evaluate the following integral: $$\int t^2 e^{-t} dt$$ Answer: $$\int t^2 e^{-t} dt = -t^2 e^{-t} + 2te^{-t} + 2e^{-t} + C$$

Step by step solution

01

Choose u and dv

Choose \(u\) and \(dv\) using the LIATE rule (Logarithm, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions). In this case, choose \(u = t^2\) (Algebraic) and \(dv = e^{-t} dt\) (Exponential).
02

Calculate du and v

Differentiate \(u\) with respect to \(t\) to get \(du\), and integrate \(dv\) with respect to \(t\) to get \(v\): $$ du = \frac{d}{dt}(t^2) dt = 2tdt $$ and $$ v = \int e^{-t} dt = -e^{-t} $$
03

Apply integration by parts formula

Now we apply the formula from our analysis: $$ \int t^2 e^{-t} dt = uv - \int v du = (-t^2 e^{-t}) - \int (-2t e^{-t})dt $$
04

Apply integration by parts again

The resulting integral still requires integration by parts: $$ u_1 = t, \quad dv_1 = -2e^{-t}dt $$ So, $$ du_1 = dt, \quad v_1 = 2e^{-t} $$ Apply the integration by parts formula again: $$ \int (-2t e^{-t})dt = u_1v_1 - \int v_1 du_1 = -2te^{-t} - \int 2e^{-t} dt = -2te^{-t} - 2e^{-t} + C $$
05

Substitute back into the original integral

Now substitute this result back into the original integral: $$ \int t^2 e^{-t} dt = -t^2 e^{-t} -\left( -2te^{-t} - 2e^{-t} \right) + C $$
06

Simplify the final expression

Finally, simplify the result by combining terms: $$ \int t^2 e^{-t} dt = -t^2 e^{-t} + 2te^{-t} + 2e^{-t} + C $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Definite Integrals
In calculus, a definite integral computes the accumulation of quantities, which can be visualized as the area under a curve. When evaluating a definite integral, you're looking for the net area under a curve within given bounds. This differs from an indefinite integral, which includes a constant of integration, represented as +C.
For example, a definite integral of a function like \(f(t)\) between two limits \(a\) and \(b\) is written as:
  • \( \int_{a}^{b} f(t) \, dt \)
Here, you'll obtain a specific numerical value, which can represent things like distance, area, or total accumulated change over an interval. Employing integration techniques like integration by parts is often necessary when integrands involve complex functions, such as the product of algebraic and exponential functions.
The Nature of Exponential Functions
Exponential functions are widely used in calculus due to their unique properties. An exponential function is typically expressed in the form \(f(t) = a e^{kt}\), where \(e\) represents Euler's number, approximately 2.718. The constant \(k\) is crucial because it determines the growth or decay of the function.
For instance, in the expression \(e^{-t}\) seen in the integration exercise, the function represents exponential decay. This is because the coefficient of \(t\) is negative, making the function decrease over time. Such characteristics make exponential functions highly applicable in modeling real-world situations, like population growth or radioactive decay.
One key feature of exponential functions is their property that differentiating or integrating them yields another exponential function, as seen in the solution where:
  • \(\frac{d}{dt} (e^{-t}) = -e^{-t}\)
  • \(\int e^{-t} dt = -e^{-t} + C\)
These properties are very useful in advanced integration techniques like integration by parts.
Using the LIATE Rule
The LIATE rule guides us in selecting which part of an integrand should be \(u\) and which part should be \(dv\) in the integration by parts technique. LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions.
The rule suggests prioritizing the choice of \(u\) according to this order, not just its type but also its presence in the formula, which helps simplify the integration process.
In the given exercise, \( t^2 \) falls under Algebraic, while \( e^{-t} \) is an Exponential function. According to LIATE:
  • Choose \(u = t^2\) because it is higher on the list (Algebraic).
  • Then set \(dv = e^{-t} dt\).
This strategic selection simplifies the computations by ensuring that differentiating \(u\) or integrating \(dv\) will lead to simpler expressions. This helps transform a complex integral into easier-to-solve components.

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Most popular questions from this chapter

Use a computer algebra system to evaluate the following definite integrals. In each case, find an exact value of the integral (obtained by a symbolic method) and find an approximate value (obtained by a numerical method). Compare the results. $$\int_{0}^{\pi / 2} \frac{d t}{1+\tan ^{2} t}$$

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Explain how to solve a separable differential equation of the form \(g(y) y^{\prime}(t)=h(t)\).

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Determine whether the following statements are true and give an explanation or counterexample. a. The Trapezoid Rule is exact when used to approximate the definite integral of a linear function. b. If the number of subintervals used in the Midpoint Rule is increased by a factor of \(3,\) the error is expected to decrease by a factor of 8. c. If the number of subintervals used in the Trapezoid Rule is increased by a factor of \(4,\) the error is expected to decrease by a factor of 16.

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