# 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 must integrate it twice to find the function. This produces two integrating constants, so we need two other facts in the curve 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} ## Mastering Integration by Parts: The Ultimate Guide

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