Test: Integrals- 1 - JEE MCQ

Acceleration is the derivative of velocity, so we integrate a(t) to get v(t):

v(t)=∫a(t)dt=∫sin(t)dt=−cos(t)+C

Now, we are given that the velocity at t=0 is 10 km/hr. We can use this information to find the constant C:

v(0)=−cos(0)+C=−1+C=10

Solving for C, we get C=11.

Now, we have the velocity function:

v(t)=−cos(t)+11

Finally, we integrate v(t) to get the displacement function s(t):

s(t)=∫v(t)dt=∫(−cos(t)+11)dt

s(t)=−sin(t)+11t+D

Now, we need to find the constant D. We are given that the car moves from t=0 to t=π/2, and we know that s(0)=0 (starting position). Plugging in these values, we can solve for D:

s(0)=−sin(0)+11(0)+D=0

D=0

So, the displacement function is:

s(t)=−sin(t)+11t

Now, to find the distance traveled, we evaluate s(t) over the given time interval:

Distance=s(π/2​)−s(0)

Distance=(−sin(π/2​)+11(π/2​))−(−sin(0)+11(0))

Distance=−1+11π​/2 = 16.27887 kilometers

Therefore, the distance traveled by the car from t=0 to t=π/2​ is 16.27887 kilometers​.

Using By parts,

2I = x|x|

I=x|x|/2

The greatest integer function ⌊x⌋ returns the largest integer less than or equal to x. For x², this means that ⌊x²⌋ will be the greatest integer less than or equal to x².

The integral becomes:

The function ⌊x²⌋ will be 0 on the interval (0,1)(0,1), 1 on the interval [1,√2​], and 4 on the interval [√2,2].

So, the integral is the sum of the areas of these intervals:

Evaluating these integrals:

8 - 1 + √2 - 4√2

7 - 3√2

So, the value of the integral is 7-3√2.

The correct option is d : 26/3

Since g(x) and h(x) are integrals of the same function , therefore ; g(x) – h(x) is constant.