Question

Evaluate the following definite integral as limit of sums:$\int ^5_0 \ {(x+1)} \, dx$

Answer 1

Answer:

The value of the given integral is

$\frac{\pi}{8}$

Step-by-step explanation:

Formula used:

$\int{\frac{1}{x^2+a^2}}\:dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+c$

Now.

$\int\limits^1_{-1}\frac{1}{x^2+2x+5}\:dx\\\\=\int\limits^1_{-1}\frac{1}{(x^2+2x+1)+4}\:dx\\\\=\int\limits^1_{-1}\frac{1}{(x+1)^2+2^2}\:dx\\\\=\frac{1}{2}[tan^{-1}\frac{x+1}{2}]^1_{-1}\\\\=\frac{1}{2}[tan^{-1}\frac{2}{2}-tan\frac{0}{2}]\\\\=\frac{1}{2}[tan^{-1}(1)-tan(0)]\\\\=\frac{1}{2}[\frac{\pi}{4}]\\\\=\frac{\pi}{8}$

Answer 2

Answer:

$\boxed{\displaystyle \int_{0}^{5}(x+1)dx=\dfrac{35}{2}}$

Step-by-step explanation:

Condition and formula for integral as limit of sums:

$\displaystyle\int _ { 0 } ^ { b } f ( x ) d x = \lim } _ { n \rightarrow \infty } h \sum _ { r = 0 } ^ { n - 1 } f ( a + r h )$

here, h = (b - a) / n

$\displaystyle \int_{0}^{5}(x+1)dx=\lim_{n\rightarrow\infty}(\frac{5}{n})\sum_{r=0}^{n-1}f(\frac{5r}{n})$

$\displaystyle=\lim } _ { n \rightarrow \infty } ( \frac { 5 } { n } ) \sum _ { r = 0 } ^ { n - 1 } ( \frac { 5 r } { n } ) + 1\\\\\= \lim } _ { n \rightarrow \infty } ( \frac { 5 } { n } ) ( \frac { 5 ( n - 1 ) ( n ) } { 2 n } + ( n - 1 ) )$

$\displaystyle=\lim_{n \rightarrow \infty}\frac { 5 } { n }\cdot\frac { 5 n ^ { 2 } - 5 n + 2 n ^ { 2 } - 2 n } { 2 n }\\=\lim _ { n \rightarrow \infty } \frac { 5 } { n } \cdot \frac { 7 n ^ { 2 } - 7 n } { 2 n }\\= \lim } _ { n \rightarrow \infty } \frac { 35 n ^ { 2 } - 35 n } { 2 n ^ { 2 } }\\=\lim } _ { n \rightarrow \infty } \frac { 35 } { 2 } - ( \frac { 35 } { 2 n } )\\=\frac { 35 } { 2 }$