Question

Find the exact solution given that f^' (x)=8x-6e^x and that f(0)=2.
This means to integrate the above function and find the function f(x)?

Answer 1

Answer:

$\text{f}(x)=4x^2-6e^x+8$

Step-by-step explanation:

Given:

• $\text{f}\:'(x)=8x-6e^x$

• $\text{f}(0)=2$

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

To find the function f(x), integrate f'(x) and use f(0) = 2 to find the value of the constant.

$\begin{aligned}\displaystyle \int (8x-6e^x)\:\:\text{d}x & = \int 8x\:\:\text{d}x -\int 6e^x\:\:\text{d}x \\\\& =8 \int x\:\:\text{d}x -6\int e^x\:\:\text{d}x \\\\& = 8 \cdot \dfrac{1}{2}x^2-6e^x+\text{C}\\\\& = 4x^2-6e^x+\text{C}\end{aligned}$

To find the value of C, substitute x = 0 into the function and set it to 2:

$\begin{aligned}\text{f}(0) & =2\\\implies 4(0)^2-6e^{0}+\text{C} & =2\\0-6+\text{C} & =2\\\text{C} & =8\end{aligned}$

Finally, substitute the found value of C into the equation:

$\text{f}(x)=4x^2-6e^x+8$

Rules of Integration

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\ \displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x \end{minipage}}$

If the terms are multiplied by constants, take them outside the integral.

$\boxed{\begin{minipage}{3.5 cm}\underline{Integrating x^n }\\\\ \displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C} \end{minipage}}$

Increase the power by 1, then divide by the new power.

$\boxed{\begin{minipage}{3.5 cm}\underline{Integration of e^{x} } \\\\ \displaystyle \int e^{x}\:\text{d}x=e^{x}+\text{C} \\\\for a\neq 0 \end{minipage}}$

Learn more about integration here:

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