Questions: Integrals
Integrals26 min readLesson 7 of 8
Q1 Define an antiderivative and explain why the indefinite integral carries a constant. Check one antiderivative by differentiation.
Answer (English): A function F is an antiderivative of f on an interval if F dash equals f at every point of that interval. If F is one antiderivative, then F plus any constant is also an antiderivative, because the derivative of a constant is zero. The indefinite integral writes the whole family as F plus C, and an extra condition on the function selects one member. As a check, the antiderivative of three x squared plus two is x cubed plus two x plus C, and differentiating returns three x squared plus two.
Hinglish: Antiderivative woh function hai jiska derivative integrand de. Constant ka derivative zero hota hai, isliye poora family F + C likhte hain. Check karne ke liye answer differentiate karo; wapas integrand aana chahiye.
Q2 Evaluate the integral of two x plus three whole cubed with respect to x.
Answer (English): Put u equal to two x plus three, so du is two dx and dx is du by two. The integral becomes one half the integral of u cubed du, which is one half times u to the fourth over four, giving u to the fourth over eight. Replacing u gives two x plus three whole to the fourth over eight plus C. Differentiating this answer returns two x plus three whole cubed times two times four over eight, which is exactly the integrand.
Hinglish: u = 2x+3 lo, to du = 2dx. Integral aadha u^3 du ban jaata hai, jiska answer u^4/8 + C hai. Wapas u ki value daalo aur differentiate karke check karo.
Q3 Evaluate the integral of x over x squared plus one.
Answer (English): The numerator is, up to a factor of two, the derivative of the denominator. Put u equal to x squared plus one, so du is two x dx, and x dx is du by two. The integral becomes half the integral of one over u du, which is half the natural logarithm of the absolute value of u. Replacing u gives half the logarithm of x squared plus one, plus C; the absolute value can be dropped because x squared plus one is always positive.
Hinglish: Denominator ka derivative 2x hai, aur numerator x. u = x^2+1 lo to du = 2x dx, isliye answer (1/2) log(x^2+1) + C aata hai.
Q4 Evaluate the integral of sine cubed x with respect to x.
Answer (English): Write sine cubed x as sine x times the quantity one minus cosine squared x. Put u equal to cosine x, so du is minus sine x dx. The integral becomes minus the integral of one minus u squared du, which is minus u plus u cubed over three plus C. Replacing u gives minus cosine x plus cosine cubed x over three plus C. Differentiating confirms the sign pattern.
Hinglish: sin^3x = sinx(1-cos^2x) likho. u = cosx lo, du = -sinx dx. Answer -cosx + cos^3x/3 + C milta hai. Sign dhyan se rakho.
Q5 Evaluate the integral of one over the product of x plus one and x plus two.
Answer (English): Split the integrand into A over x plus one plus B over x plus two. Clearing the denominator gives one equal to A times x plus two plus B times x plus one. Putting x equal to minus one gives A equal to one, and putting x equal to minus two gives B equal to minus one. The integral is therefore the logarithm of x plus one minus the logarithm of x plus two, which is the logarithm of the ratio of x plus one to x plus two, plus C.
Hinglish: Partial fractions se 1/((x+1)(x+2)) = 1/(x+1) - 1/(x+2) likho, aur integral log|x+1| - log|x+2| + C aa jaata hai.
Q6 Evaluate the integral of x cosine x with respect to x using by parts.
Answer (English): Take u equal to x and v prime equal to cosine x, so v is sine x. The integral is x sine x minus the integral of sine x, which is x sine x plus cosine x plus C. The LIATE order chose the algebraic factor as u because its derivative is simpler. Differentiating the answer gives sine x plus x cosine x minus sine x, which is x cosine x, as required.
Hinglish: u = x aur v' = cosx lo. Answer x sinx + cosx + C. LIATE ke hisaab se algebraic factor u banta hai kyunki uska derivative aasan hota hai.
Q7 Evaluate the integral of one over x squared plus four x plus eight.
Answer (English): Complete the square in the denominator: x squared plus four x plus eight is x plus two whole squared plus four. The integral is now one over u squared plus two squared with u equal to x plus two and a equal to two. The standard form gives one over two times the inverse tangent of u over two, so the answer is half the inverse tangent of x plus two over two, plus C.
Hinglish: Denominator ko (x+2)^2 + 4 likho. Standard form se answer (1/2) tan^-1((x+2)/2) + C hai. Square complete karna pehla step hai.
Q8 Evaluate the definite integral from zero to one of x squared over one plus x cubed.
Answer (English): Put u equal to one plus x cubed, so du is three x squared dx. When x is zero, u is one, and when x is one, u is two. The integral becomes one third of the integral from one to two of one over u du, which is one third of the logarithm of two. The limits were changed with the substitution, so no back substitution is needed.
Hinglish: u = 1+x^3 lo, du = 3x^2 dx. Limits 0 aur 1 se u ke limits 1 aur 2 ban jaate hain. Answer (1/3) log 2 aata hai. Limits badalna zaroori hai.
Q9 Using a property of definite integrals, evaluate the integral from zero to pi by two of sine x over sine x plus cosine x.
Answer (English): Let the integral be I. Apply the reflective property, replacing x by pi by two minus x. The integrand becomes cosine x over cosine x plus sine x, which is the complement of the original fraction. Adding the two expressions for I gives twice I equal to the integral of one over the interval, whose value is pi by two. Therefore I is pi by four. No antiderivative was needed, which is the point of the property.
Hinglish: Reflective property lagao: x ko pi/2 - x se replace karo. Dono integrals add karo to 2I = pi/2 milta hai, isliye I = pi/4. Is sawaal mein antiderivative ki zaroorat hi nahi padti.
Q10 Evaluate the integral from zero to one of x plus one as the limit of a sum, and verify the answer with the fundamental theorem.
Answer (English): Divide the interval into n equal parts with h equal to one over n and evaluate at the right end points: the sum is h times the sum over r of (r over n plus one). The first sum is n(n+1) over two and the second is n, so the expression becomes (n+1) over 2n plus one. As n tends to infinity this tends to one half plus one, which is three halves. The antiderivative is x squared over two plus x, and between zero and one it gives one half plus one, also three halves, so the two methods agree.
Hinglish: h = 1/n, right end points par sum lo. Sum simplify karke (n+1)/2n + 1 milta hai, jiska limit 3/2 hai. Fundamental theorem se bhi 3/2 aata hai, isliye answer verify ho jaata hai.
The chapter quiz on this page has five questions on the standard forms, substitution, by parts, definite integral properties and the limit of a sum. Attempt it after writing the ten answers above.