45. solve for $x$: $int_{0}^{x}(t^{3}-2t + 3)dt = 4$.

Explanation:

Step1: Integrate the function

We know that $\int(t^{3}-2t + 3)dt=\frac{t^{4}}{4}-t^{2}+3t + C$. Then $\int_{0}^{x}(t^{3}-2t + 3)dt=\left[\frac{t^{4}}{4}-t^{2}+3t\right]_{0}^{x}=\frac{x^{4}}{4}-x^{2}+3x$.

Step2: Set up the equation

Set $\frac{x^{4}}{4}-x^{2}+3x = 4$, or $x^{4}-4x^{2}+12x - 16=0$. We can try some simple values of $x$. When $x = 2$: $2^{4}-4\times2^{2}+12\times2 - 16=16-16 + 24-16=8\neq0$. When $x = 1$: $1^{4}-4\times1^{2}+12\times1 - 16=1-4 + 12-16=-7\neq0$. We can use a numerical - method like Newton - Raphson method. Let $f(x)=x^{4}-4x^{2}+12x - 16$, then $f^\prime(x)=4x^{3}-8x + 12$. The Newton - Raphson formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^\prime(x_{n})}$. Starting with an initial guess $x_{0}=1$: $f(1)=1 - 4+12 - 16=-7$, $f^\prime(1)=4-8 + 12 = 8$. $x_{1}=1-\frac{-7}{8}=\frac{15}{8}=1.875$. $f(1.875)=(1.875)^{4}-4\times(1.875)^{2}+12\times1.875 - 16\approx11.39-14.06+22.5 - 16\approx3.83$. $f^\prime(1.875)=4\times(1.875)^{3}-8\times1.875 + 12\approx25.39-15+12\approx22.39$. $x_{2}=1.875-\frac{3.83}{22.39}\approx1.875 - 0.171\approx1.704$. Continuing the iterations, we find that $x\approx1.67$.

Answer:

$x\approx1.67$