∫14x+√x+1/xdx-∫12log2dx=

Correct answer: (A) 79/6. Let I=∫14x+√x+1/xdx-∫12log2dx ⇒I=x^22+2/3x^32+logx14-x12log2 ⇒I=16/2+(2×8)/3+log4-1^22+2/3+log1-2log2-1 ⇒I=15/2+14/3+2log2-2log2-1 ⇒I=79/6

Q116 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Mathematics

$\int_{1}^{4} (x + \sqrt{x} + \frac{1}{x}) d x - \int_{1}^{2 \log 2} d x =$

Correct answer
A
$\frac{79}{6}$
B
$\frac{643}{6}$
C
$\frac{321}{5}$
D
$64$

Solution

Let
$I = \int_{1}^{4} (x + \sqrt{x} + \frac{1}{x}) d x - \int_{1}^{2 \log 2} d x$
$\Rightarrow I = {[\frac{x^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} + \log x]}_{1}^{4} - {[x]}_{1}^{2 \log 2}$
$\Rightarrow I = [(\frac{16}{2} + \frac{2 \times 8}{3} + \log 4) - (\frac{1^{2}}{2} + \frac{2}{3} + \log 1)] - [2 \log 2 - 1]$
$\Rightarrow I = (\frac{15}{2} + \frac{14}{3} + 2 \log 2) - [2 \log 2 - 1]$
$\Rightarrow I = \frac{79}{6}$

Hence, the correct answer is option: 'A'.

∫14x+x+1xdx-∫12log2dx=

Solution

Relevant Questions

$\int_{1}^{4} (x + \sqrt{x} + \frac{1}{x}) d x - \int_{1}^{2 \log 2} d x =$