Particular Values of Integration


We can find the integral constant $c$, given a particular value of the function.

Example 1

Find $f(x)$ given that $f'(x) = 3x^2+4x-5$ and $f(1)=3$.

\( \begin{align} \displaystyle
f(x) &= \int{(3x^2+4x-5)}dx \\
&= \dfrac{3x^{2+1}}{2+1} + \dfrac{4x^{1+1}}{1+1} – 5x +c \\
&= x^3 + 2x^2 – 5x + c \\
f(1) &= 3 \\
1^3 + 2 \times 1^2 – 5 \times 1 + c &= 3 \\
c &= 5 \\
\therefore f(x) &=x^3 + 2x^2 – 5x + 5 \\
\end{align} \)

If we are given the second derivative we need to integrate twice to find the function. This produces two integrating constants, so we need two other facts the curve in order to determine these constants.

Example 2

Find $f(x)$ given that $f^{\prime \prime}(x) = 12x^2-4$ and $f'(0)=2$ and $f(1)=4$.

\( \begin{align} \displaystyle
f'(x) &= \int{(12x^2-4)}dx \\
&= \dfrac{12x^{2+1}}{2+1} – 4x + c \\
&= 4x^3 -4x + c \\
f'(0) &= 2 \\
4 \times 0^3 – 4 \times 0 + c &= 2 \\
c &= 2 \\
f'(x) = 4x^3 -4x + 2 \\
f(x) &= \dfrac{4x^{3+1}}{3+1} – \dfrac{4x^{1+1}}{1+1} + 2x + d \\
&= x^4 – 2x^2 + 2x + d \\
f(1) &= 4 \\
1^4 – 2 \times 1^2 + 2 \times 1 + d &= 4 \\
d &= 3 \\
\therefore f(x) &= x^4 – 2x^2 + 2x + 3 \\
\end{align} \)


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