MA005: Calculus I

Practice 1: $\begin{aligned} &A(1)=1, A(2)=1.5, A(3)=1, A(4)=0.5 \\ &A^{\prime}(x)=f(x) \text { so } A^{\prime}(1)=f(1)=1, A^{\prime}(2)=f(2)=0, A^{\prime}(3)=-1, A^{\prime}(4)=0 \end{aligned}$

Practice 2: $F(x)=x^{3}-x$ is one antiderivative of $f(x)=3 x^{2}-1\left(F^{\prime}=f\right)$ so

$\int_{1}^{3} 3 x^{2}-1 \mathrm{dx}=\mathrm{x}^{3}-\left.\mathrm{x}\right|_{1} ^{3}=\left(3^{3}-3\right)-\left(1^{3}-1\right)=24$.

$F(x)=x^{3}-x+7$ is another antiderivative of $f(x)=3 x^{2}-1$ so

$\int_{1}^{3} 3 x^{2}-1 d x=x^{3}-x+\left.7\right|_{1} ^{3}=\left(3^{3}-3+7\right)-\left(1^{3}-1+7\right)=24$

No matter which antiderivative of $f(x)=3 x^{2}-1$ you use, the value of the definite integral $\int_{1}^{3} 3 x^{2}-1 d x \text { is 24. }$

Practice 3: $\int_{1.3}^{3.4} \operatorname{INT}(x) d x=3.9$. Since $f(x)=\operatorname{INT}(x)$ is not continuous on the interval [1.3, 3.4] so we can not use the Fundamental Theorem of Calculus. Instead, we can think of the definite integral as an area (Fig. 20).

Practice 4: Total volume $\lim _{m e s h \rightarrow 0}\left\{\sum_{i=1}^{N}\left(3-c_{i}\right) \cdot\left(3-c_{i}\right) \cdot \Delta \mathrm{x}_{\mathrm{i}}\right\}$.

$\begin{aligned} &=\int_{0}^{2}(3-x)(3-x) d x \\ &=\int_{0}^{2}\left(9-6 x+x^{2}\right) d x \\ &=9 x-3 x^{2}+\left.\frac{x^{3}}{3}\right|_{0} ^{2}=\left(18-12+\frac{8}{3}\right)-(0-0+0)=8 \frac{2}{3} \end{aligned}$

Practice 5: $\frac{d}{d x}\left(\int_{0}^{x^{3}} \sin (t) d t\right)=\sin \left(x^{3}\right) \frac{d x^{3}}{d x}=3 x^{2} \cdot \sin \left(x^{3}\right)$.