By part 2 of the Fundamental Theorem of Calculus, $\int _ { 0 } ^ { 1 } e ^ { x } d x$ is

A) $e ^ { 2 } - 1$
B) $e ^ { 2 }$
C) $e + 1$
D) $e$
E) $e - 1$

Question

By part 2 of the Fundamental Theorem of Calculus, $\int _ { 0 } ^ { 1 } e ^ { x } d x$ is

A) $e ^ { 2 } - 1$
B) $e ^ { 2 }$
C) $e + 1$
D) $e$
E) $e - 1$

Quizplus · Accepted Answer

The Fundamental Theorem of Calculus Part 2 states that if $F$ is an antiderivative of $f$ on an interval $[a, b]$, then $\int _ { a } ^ { b } f(x) dx = F(b) - F(a)$. For $f(x) = e^x$, an antiderivative is $F(x) = e^x$. Thus, $\int _ { 0 } ^ { 1 } e ^ { x } dx = e^1 - e^0 = e - 1$.

By Part 2 of the Fundamental Theorem of Calculus ∫01exdx\int _ { 0 } ^ { 1 } e ^ { x } d x∫01​exdx

By Part 2 of the Fundamental Theorem of Calculus $\int _ { 0 } ^ { 1 } e ^ { x } d x$