Fundamental Theorem of Calculus Part 2
If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$, then
$
\int_a^b f(x) d x=F(b)-F(a) .
$
Example1
- $\int_{\pi / 2}^x \cos t d t=$
▶️Answer/Explanation
Solution
$
\text { } \int_{\pi / 2}^x \cos t d t=[\sin t]_{\pi / 2}^x=\left[\sin x-\sin \frac{\pi}{2}\right]=\sin x-1
$
Example 2
- The graph of the function $f$ shown below consists of four line segments. If $g$ is the function defined by $g(x)=\int_{-4}^x f(t) d t$, find the value of $g(6), g^{\prime}(6)$, and $g^{\prime \prime}(6)$.
▶️Answer/Explanation
Solution
$\begin{array}{rlrl}g(6)= & \int_{-4}^6 f(t) d t & & \text { Substitute } 6 \text { for } x . \\ = & \frac{1}{2}(2)(4)-\frac{1}{2}(2)(2)+\frac{1}{2}(2)(4) & & \int_{-4}^6 f(t) d t=\text { sum of the area above } \\ & +(3)(4)+\frac{1}{2}(2+4) \cdot 1 & & \text { the } x \text {-axis minus sum of the area belo } \\ = & 21 & & \text { theaxis, between } x=-4 \text { and } x=6 . \\ g^{\prime}(x) & =f(x) & & \text { Fundamental Theorem of Calculus } \\ g^{\prime}(6) & =f(6)=2 & & \text { Slope of } f \text { at } x=6 \text { is }-2 . \\ g^{\prime \prime}(6) & =f^{\prime}(6)=-2 & \end{array}$
Example 3
- $f$ is the antiderivative of $\frac{\sqrt{x}}{1+x^3}$ such that $f(1)=2$, then $f(3)=$
(A) 1.845
(B) 2.397
(C) 2.906
(D) 3.234
▶️Answer/Explanation
Ans:B
Example 4
- If $f^{\prime}(x)=\cos \left(x^2-1\right)$ and $f(-1)=1.5$, then $f(5)=$
(A) 1.554
(B) 2.841
(C) 3.024
(D) 3.456
▶️Answer/Explanation
Ans:C
Example 5
- If $f(x)=\sqrt{x^4-3 x+4}$ and $g$ is the antiderivative of $f$, such that $g(3)=7$, then $g(0)=$
(A) -2.966
(B) -1.472
(C) -0.745
(D) 1.086
▶️Answer/Explanation
Ans:A
Example 6
- If $f$ is a continuous function and $F^{\prime}(x)=f(x)$ for all real numbers $x$, then $\int_2^{10} f\left(\frac{1}{2} x\right) d x=$
(A) $\frac{1}{2}[F(5)-F(1)]$
(B) $\frac{1}{2}[F(10)-F(2)]$
(C) $2[F(5)-F(1)]$
(D) $2[F(10)-F(2)]$
▶️Answer/Explanation
Ans:C