The Fundamental Theorem of Calculus

Read this section to see the connection between derivatives and integrals. Work through practice problems 1-5.

Practice Answers

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

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\}.

    &=\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}

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).