Q.Evaluate the following definite integral:
$\int_1^3 \log x\, dx$

$\displaystyle\int_0^a 3x^2\,dx = a^3 = 8$, so $a = 2$ — option (i).

Integrate and apply the limits using the fundamental theorem of calculus:

$\int_0^a 3x^2\,dx = \left[\,x^3\,\right]_0^a = a^3 - 0 = a^3.$

…

Since $\int e^x\,dx = e^x$, evaluating between $0$ and $2$ gives $e^2 - 1$.

By the fundamental theorem of calculus, $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$, where $F$ is an antiderivative of $f$. Here $f(x) = e^x$ and its antiderivative is $F(x) = e^x$.

Evaluate at the limits: …