retagged by
275 views
0 votes
0 votes

The area under the curve $x^2+3x-4$ in the positive quadrant and bounded by the line $x=5$ is equal to

  1. $59 \frac{1}{6}$
  2. $61 \frac{1}{3}$
  3. $40 \frac{2}{3}$
  4. $72$
retagged by

1 Answer

0 votes
0 votes
The function  $y = x^{2} + 3x - 4$ intersects the x-axis at $\left ( 1,0 \right )$

We will take only first quadrant into consideration. So, the intersection point of the given equation and the line $x = 5$  is the point $\left ( 5,36 \right )$. (put $x = 5$ in the quadratic equation and $y$ turns out be 36).

So, the answer to this question is the area under the curve of the given equation and the line  $x = 5$  i.e. from $x = 1$  to $x = 5$ .

This can easily be calculated using integration.

 

$\int_{1}^{5} x^{2} + 3x - 4$   $dx$   = $\frac{368}{6}$   =  $61 \frac{1}{3}$  

Option B is the correct answer.
edited by

Related questions

0 votes
0 votes
1 answer
1
gatecse asked Sep 18, 2019
324 views
The area bounded by $y=x^{2}-4,y=0$ and $x=4$ is$\frac{64}{3}$$6$$\frac{16}{3}$$\frac{32}{3}$
0 votes
0 votes
0 answers
2
Arjun asked Sep 23, 2019
310 views
Let $y^2-4ax+4a=0$ and $x^2+y^2-2(1+a)x+1+2a-3a^2=0$ be two curves. State which one of the following statements is true.These two curves intersect at two pointsThese two ...
0 votes
0 votes
1 answer
3
Arjun asked Sep 23, 2019
322 views
If $A(t)$ is the area of the region bounded by the curve $y=e^{-\mid x \mid}$ and the portion of the $x$-axis between $-t$ and $t$, then $\underset{t \to \infty}{\lim} A(...
0 votes
0 votes
0 answers
4
gatecse asked Sep 18, 2019
362 views
The shaded region in the following diagram represents the relation$y\:\leq\: x$$\mid \:y\mid \:\leq\: \mid x\:\mid $$y\:\leq\: \mid x\:\mid$$\mid \:y\mid\: \leq\: x$