11th Biology

Evaluation of the Definite Integral \( \int_0^1 \sin(x^2+2) \,dx \) using the Gaussian Quadrature Formula with Four Points

To evaluate the definite integral \( \int_0^1 \sin(x^2+2) \,dx \) using the Gaussian quadrature formula with four points, we first need to transform the interval of integration [0,1] to the standard interval [-1,1].

Let \( x=\frac{1}{2}(t+1) \), then \( dx=\frac{1}{2}dt \). When \( x=0 \), we get \( t=-1 \); and when \( x=1 \), we get \( t=1 \).

The integral becomes:

\[ \int_0^1 \sin(x^2+2) \,dx = \int_{-1}^1 \sin\left(\left(\frac{t+1}{2}\right)^2+2\right)\frac{1}{2}dt \]

Let \( f(t)=\frac{1}{2}\sin\left(\left(\frac{t+1}{2}\right)^2+2\right) \).

Gaussian Quadrature Formula

The Gaussian quadrature formula with four points is given by:

\[ \int_{-1}^1 f(t) \,dt \approx \sum_{i=1}^4 w_i f(t_i) \]

Weights and Points

The weights \( w_i \) and points \( t_i \) for a four-point Gaussian quadrature are:

\[ t_1 = -\sqrt{\frac{3}{7} + \frac{2}{7}\sqrt{\frac{6}{5}}} \approx -0.86113631 \] \[ t_2 = -\sqrt{\frac{3}{7} - \frac{2}{7}\sqrt{\frac{6}{5}}} \approx -0.33998104 \] \[ t_3 = \sqrt{\frac{3}{7} - \frac{2}{7}\sqrt{\frac{6}{5}}} \approx 0.33998104 \] \[ t_4 = \sqrt{\frac{3}{7} + \frac{2}{7}\sqrt{\frac{6}{5}}} \approx 0.86113631 \] \[ w_1 = w_4 = \frac{18 - \sqrt{30}}{36} \approx 0.34785485 \] \[ w_2 = w_3 = \frac{18 + \sqrt{30}}{36} \approx 0.65214515 \]

Evaluation at Each Point

Next, we evaluate \( f(t) \) at each of these points:

For \( t_1 = -0.86113631 \):

\[ x_1 = \frac{1}{2}(t_1+1) = \frac{1}{2}(-0.86113631+1) = 0.069431845 \] \[ f(t_1) = \frac{1}{2}\sin((0.069431845)^2+2) = \frac{1}{2}\sin(2.0048201) = \frac{1}{2}(0.9056636) = 0.4528318 \]

For \( t_2 = -0.33998104 \):

\[ x_2 = \frac{1}{2}(t_2+1) = \frac{1}{2}(-0.33998104+1) = 0.33000948 \] \[ f(t_2) = \frac{1}{2}\sin((0.33000948)^2+2) = \frac{1}{2}\sin(2.108929) = \frac{1}{2}(0.8354078) = 0.4177039 \]

For \( t_3 = 0.33998104 \):

\[ x_3 = \frac{1}{2}(t_3+1) = \frac{1}{2}(0.33998104+1) = 0.66999052 \] \[ f(t_3) = \frac{1}{2}\sin((0.66999052)^2+2) = \frac{1}{2}\sin(2.448887) = \frac{1}{2}(0.6508922) = 0.3254461 \]

For \( t_4 = 0.86113631 \):

\[ x_4 = \frac{1}{2}(t_4+1) = \frac{1}{2}(0.86113631+1) = 0.930568155 \] \[ f(t_4) = \frac{1}{2}\sin((0.930568155)^2+2) = \frac{1}{2}\sin(2.866057) = \frac{1}{2}(0.2764515) = 0.13822575 \]

Final Calculation

Finally, we apply the Gaussian quadrature formula:

\[ \int_0^1 \sin(x^2+2) \,dx \approx \sum_{i=1}^4 w_i f(t_i) \] \[ = w_1 f(t_1) + w_2 f(t_2) + w_3 f(t_3) + w_4 f(t_4) \] \[ = (0.34785485)(0.4528318)+(0.65214515)(0.4177039)+(0.65214515)(0.3254461)+(0.34785485)(0.13822575) \] \[ = 0.157523+0.272494+0.212217+0.048106 \] \[ \approx 0.69034 \]

Therefore, the approximate value of the integral is \( 0.69034 \).