Fundamental Theorem of Calculus Part 1
If $f$ is continuous on $[a, b]$ then $F(x)=\int_a^x f(t) d t$ is continuous on $[a, b]$ and differentiable on $(a, b)$, so
$
F^{\prime}(x)=\frac{d}{d x} \int_a^x f(t) d t=f(x)
$
If $F(x)=\int_a^{u(x)} f(t) d t$, then by the chain rule
$
F^{\prime}(x)=\frac{d}{d x} \int_a^{u(x)} f(t) d t=f(u) \cdot \frac{d u}{d x}
$
Example1
- If $F(x)=\int_1^x \frac{1}{1+t^3} d t$, then $F^{\prime}(x)=$
▶️Answer/Explanation
Solution
$
\begin{aligned}
& F^{\prime}(x)=\frac{d}{d x} \int_1^x \frac{1}{1+t^3} d t \\
& =\frac{1}{1+x^3}
\end{aligned}
$
Fundamental Theorem of Calculus
Example2
- $
\frac{d}{d x} \int_1^{x^2} \sqrt{3+t^2} d t=
$
(A) $\sqrt{3+x^2}$
(B) $\sqrt{3+x^4}$
(C) $2 x \sqrt{3+x^4}$
(D) $2 \sqrt{3+x^2}$
▶️Answer/Explanation
Ans:C
Example3
- For $-\frac{\pi}{2}<x<\frac{\pi}{2}$, if $F(x)=\int_0^{\sin x} \frac{d t}{\sqrt{1-t^2}}$, then $F^{\prime}(x)=$
(A) $\frac{\sin x}{\sqrt{1-x^2}}$
(B) $\frac{\cos x}{\sqrt{1-x^2}}$
(C) 1
(D) $\csc x$
▶️Answer/Explanation
Ans:C
Example4
- If $F(x)=\int_0^{\sqrt{x}} \cos \left(t^2\right) d t$, then $F^{\prime}(4)=$
(A) $\cos 2$
(B) $\frac{\cos 4}{4}$
(C) $\frac{\cos 4}{\sqrt{2}}$
(D) $\sqrt{2} \cos 4$
▶️Answer/Explanation
Ans:B