Integrals- 1 - Free MCQ Practice Test with solutions, JEE Maths

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​.

We want to find the indefinite integral of |x| with respect to x:

∫ |x| dx

1. Consider x ≥ 0:

   • Here, |x| = x.
   • So, ∫ |x| dx = ∫ x dx = (x²)/2 + C₁

2. Consider x < 0:

   • Here, |x| = -x.
   • So, ∫ |x| dx = ∫ (-x) dx = - (x²)/2 + C₂

To combine these results into a single expression, we note that for an indefinite integral, constants C₁ and C₂ can be absorbed into a single arbitrary constant C. A concise way to write this is:

∫ |x| dx = (x · |x|)/2 + C

• For x ≥ 0, this becomes x·x/2 = x²/2.
• For x < 0, this becomes x·(-x)/2 = -x²/2.
• Combining both the results , we get x|x|/2 + C

√2 - 1 + 2√3 - 2√2 + 6 - 3√3
5 - √2 - √3

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

Let t = log x (natural log). Then 

Hence, Option (b) is the correct answer.