Evaluate ∫ 0 2 ( E 0.5 X + 2 X ) D X \int_0^2 \left(e^{0.5 X} + 2x\right) \, Dx ∫ 0 2 ​ ( E 0.5 X + 2 X ) D X And Express The Answer In Simplest Form.

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Introduction


In this article, we will evaluate the definite integral of the function (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22. This involves finding the antiderivative of the function and then applying the fundamental theorem of calculus to obtain the definite integral.

Step 1: Find the Antiderivative of e0.5xe^{0.5 x}


The antiderivative of e0.5xe^{0.5 x} is given by 20.5e0.5x=4e0.5x\frac{2}{0.5} e^{0.5 x} = 4e^{0.5 x}.

Step 2: Find the Antiderivative of 2x2x


The antiderivative of 2x2x is given by x2x^2.

Step 3: Combine the Antiderivatives


The antiderivative of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) is given by 4e0.5x+x24e^{0.5 x} + x^2.

Step 4: Apply the Fundamental Theorem of Calculus


The fundamental theorem of calculus states that the definite integral of a function f(x)f(x) from aa to bb is given by F(b)F(a)F(b) - F(a), where F(x)F(x) is the antiderivative of f(x)f(x).

Step 5: Evaluate the Definite Integral


We need to evaluate the definite integral of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22. This involves substituting the values of xx into the antiderivative and then subtracting the value of the antiderivative at x=0x=0 from the value of the antiderivative at x=2x=2.

Step 6: Calculate the Value of the Antiderivative at x=2x=2


The value of the antiderivative at x=2x=2 is given by 4e0.5(2)+(2)2=4e1+44e^{0.5(2)} + (2)^2 = 4e^1 + 4.

Step 7: Calculate the Value of the Antiderivative at x=0x=0


The value of the antiderivative at x=0x=0 is given by 4e0.5(0)+(0)2=4e0+0=44e^{0.5(0)} + (0)^2 = 4e^0 + 0 = 4.

Step 8: Subtract the Value of the Antiderivative at x=0x=0 from the Value of the Antiderivative at x=2x=2


The value of the definite integral is given by 4e1+44=4e14e^1 + 4 - 4 = 4e^1.

Step 9: Simplify the Answer


The value of the definite integral is given by 4e1=4e1=42.7182810.873124e^1 = 4e^1 = 4 \cdot 2.71828 \approx 10.87312.

Conclusion


In this article, we evaluated the definite integral of the function (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22 and expressed the answer in simplest form. The value of the definite integral is given by 4e1=42.7182810.873124e^1 = 4 \cdot 2.71828 \approx 10.87312.

Final Answer


The final answer is 4e1\boxed{4e^1}.

References


  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart

Tags


  • Calculus
  • Definite Integral
  • Antiderivative
  • Fundamental Theorem of Calculus
  • Exponential Function
  • Trigonometric Function

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Frequently Asked Questions


Q: What is the definite integral of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22?

A: The definite integral of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22 is given by 4e1=42.7182810.873124e^1 = 4 \cdot 2.71828 \approx 10.87312.

Q: How do I find the antiderivative of e0.5xe^{0.5 x}?

A: The antiderivative of e0.5xe^{0.5 x} is given by 20.5e0.5x=4e0.5x\frac{2}{0.5} e^{0.5 x} = 4e^{0.5 x}.

Q: How do I find the antiderivative of 2x2x?

A: The antiderivative of 2x2x is given by x2x^2.

Q: How do I combine the antiderivatives of e0.5xe^{0.5 x} and 2x2x?

A: The antiderivative of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) is given by 4e0.5x+x24e^{0.5 x} + x^2.

Q: How do I apply the fundamental theorem of calculus to evaluate the definite integral?

A: The fundamental theorem of calculus states that the definite integral of a function f(x)f(x) from aa to bb is given by F(b)F(a)F(b) - F(a), where F(x)F(x) is the antiderivative of f(x)f(x).

Q: How do I evaluate the definite integral of (e0.5x+2x)\left(e^{0.5 x} + 2x\right) from 00 to 22?

A: We need to substitute the values of xx into the antiderivative and then subtract the value of the antiderivative at x=0x=0 from the value of the antiderivative at x=2x=2.

Q: What is the value of the antiderivative at x=2x=2?

A: The value of the antiderivative at x=2x=2 is given by 4e0.5(2)+(2)2=4e1+44e^{0.5(2)} + (2)^2 = 4e^1 + 4.

Q: What is the value of the antiderivative at x=0x=0?

A: The value of the antiderivative at x=0x=0 is given by 4e0.5(0)+(0)2=4e0+0=44e^{0.5(0)} + (0)^2 = 4e^0 + 0 = 4.

Q: How do I simplify the answer?

A: The value of the definite integral is given by 4e1=42.7182810.873124e^1 = 4 \cdot 2.71828 \approx 10.87312.

Additional Resources


  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Khan Academy: Calculus
  • [4] MIT OpenCourseWare: Calculus

Tags


  • Calculus
  • Definite Integral
  • Antiderivative
  • Fundamental Theorem of Calculus
  • Exponential Function
  • Trigonometric Function
  • Math
  • Science
  • Education