Estimation of Definite Integral

For the integral ${\int }_0^{\pi /2}\left(8+4\cos x\right)dx$, the absolute percentage error in numerical evaluation with the Trapezoidal rule, using only the endpoints, is _____ 

(round off to one decimal place0.

For the data,

$\begin{array}{cc}x:& 0 1 2\\ f\left(x\right):& 8 5 6\end{array}$

the value of ${\int }_0^2{\left[f\left(x\right)\right]}^{2}dx$ by Trapezoidal rule will be:

Considering four subintervals, the value of ${\int }_0^1\frac{1}{1+x}dx$ by Trapezoidal rule is:

Trapezoidal rule for the evaluation of ${\int }_a^{b}f\left(x\right)dx$ requires the interval $\left[a,b\right]$ to be divided into:

Consider the below data:

$\begin{array}{ccccc}x& :& 0& 1& 2\\ f\left(x\right)& :& 4& 3& 12\end{array}$

The value of ${\int }_0^{2}f\left(x\right)dx$ by Trapezoidal rule will be:

The integral ${\int }_1^{10}{x}^{3}dx$ is approximately evaluated by Trapezoidal rule ${\int }_1^{10}{x}^3 dx$ $=3\left[\frac{1+{10}^3}{2}+\alpha +7^3\right]$ for$n=3$, then the value of $\alpha$ is

With the help of Trapezoidal rule for numerical integration and the following table

the value of ${\int }_0^{1}f\left(x\right)dx$ is

Calculate by Trapezoidal rule an approximate value of ${\int }_{-3}^3{x}^{4}dx$ by taking seven equidistant ordinates

The value of ${\int }_0^{1}xdx$ by Trapezoidal rule taking $x=4$ is

Using trapezoidal rule and taking $n=4$, the value of ${\int }_0^2\frac{dx}{1+x}$ will be

The value of $f\left(x\right)$ is given only at $x=0,\frac{1}{3},\frac{2}{3},1$. Which of the following can be used to evaluate${\int }_0^{1}f\left(x\right)dx$ approximately?

If $I={\int }_{x_0}^{x_0+nh}ydx,$ then by Trapezoidal rule $I$ is equal to 

For $n=4$, using trapezoidal rule, the value of ${\int }_0^2\frac{dx}{1+x}$ will be

Cotyledons are also called-

Let $f :\left[0, 1\right]\to \left[0, \infty \right]$ be a continuous function such that ${\int }_0^{1}f\left(x\right)dx=10.$ Which of the following statement is NOT necessarily true?

If $I_1={\int }_0^1{2}^{x^2}dx, I_2={\int }_0^1{2}^{x^3}dx, I_3={\int }_1^2{2}^{x^2}dx, I_4={\int }_1^2{2}^{x^3}dx$ then

$\text{I}\text{f}{\mathit{\text{I}}}_1=\underset{\text{0}}{\overset{\text{1}}{\int }}{2}^{{\text{x}}^2}\mathit{\text{dx}}\text{,}{\mathit{\text{I}}}_2=\underset{\text{0}}{\overset{\text{1}}{\int }}{2}^{{\text{x}}^{\text{3}}}\mathit{\text{dx}}\text{,}{\mathit{\text{I}}}_{\text{3}}=\underset{\text{1}}{\overset{\text{2}}{\int }}{2}^{{\text{x}}^2}\mathit{\text{dx}}\text{and}{\mathit{\text{I}}}_{\text{4}}=\underset{\text{1}}{\overset{\text{2}}{\int }}{2}^{{\text{x}}^3}\mathit{\text{dx}}\text{then}$