Where Mindful Learning Begins
Where Mindful Learning Begins
Chapter 7 Ex 7.8 covers evaluating definite integrals — all 22 questions solved step by step using the Second Fundamental Theorem of Calculus: find an anti derivative, then subtract its value at the lower limit from its value at the upper limit.
This exercise is where all the integration formulae from Ex 7.1–7.7 finally get applied with actual numbers attached. The method never changes — integrate first, ignore the constant of integration since it cancels out, then evaluate the antiderivative at the upper limit and subtract its value at the lower limit — but the integrands themselves cover the full range of forms you've built up so far: polynomials, trigonometric functions, exponentials, and standard rational forms. Because the technique is consistent throughout, this is a good exercise for building calculation speed and accuracy rather than learning anything new. The last two questions are MCQs on the same theme.
Anti derivative: \dfrac{x^2}{2}+x. Evaluate: \left[\dfrac{1}{2}+1\right]-\left[\dfrac{1}{2}-1\right]=\dfrac32-\left(-\dfrac12\right).
Anti derivative: \log x. Evaluate: \log3-\log2.
Anti derivative: x^4-\dfrac{5x^3}{3}+3x^2+9x. At x=2: \dfrac{98}{3}. At x=1: \dfrac{34}{3}.
Anti derivative: -\dfrac{\cos2x}{2}. Evaluate: 0-\left(-\dfrac12\right).
Anti derivative: \dfrac{\sin2x}{2}. Both \sin\pi and \sin0 are zero.
Anti derivative: e^x.
Anti derivative: \log|\sec x|. Evaluate: \log\sqrt2-\log1.
Anti derivative: \log\left|\text{cosec}\,x-\cot x\right|. At x=\dfrac{\pi}{4}: \log(\sqrt2-1). At x=\dfrac{\pi}{6}: \log(2-\sqrt3).
Anti derivative: \sin^{-1}x. Evaluate: \dfrac{\pi}{2}-0.
Anti derivative: \tan^{-1}x. Evaluate: \dfrac{\pi}{4}-0.
Anti derivative: \dfrac12\log\left|\dfrac{x-1}{x+1}\right| (standard formula with a=1).
At x=3: \dfrac12\log\dfrac12. At x=2: \dfrac12\log\dfrac13.
Use \cos^2x=\dfrac{1+\cos2x}{2}. Anti derivative: \dfrac{x}{2}+\dfrac{\sin2x}{4}; the sine term vanishes at both limits.
Anti derivative: \dfrac12\log(x^2+1). Evaluate: \dfrac12\log10-\dfrac12\log5.
Split into \dfrac{2x}{5x^2+1}+\dfrac{3}{5x^2+1}, with anti derivatives \dfrac15\log(5x^2+1) and \dfrac{3}{\sqrt5}\tan^{-1}(\sqrt5\,x). Both terms vanish at x=0.
Put t=x^2; anti derivative is \dfrac12 e^{x^2}.
Improper — divide: \dfrac{5x^2}{x^2+4x+3} = 5 - \dfrac{20x+15}{(x+1)(x+3)}. Decomposing the fraction gives 5+\dfrac{5}{2(x+1)}-\dfrac{45}{2(x+3)}.
Anti derivative: 5x+\dfrac52\log|x+1|-\dfrac{45}{2}\log|x+3|. Evaluate at x=2 and x=1 and simplify.
Anti derivative: 2\tan x+\dfrac{x^4}{4}+2x. At x=\dfrac{\pi}{4}: 2+\dfrac{\pi^4}{1024}+\dfrac{\pi}{2}; the lower limit contributes 0.
Since \sin^2\dfrac{x}{2}-\cos^2\dfrac{x}{2}=-\cos x, the anti derivative is -\sin x, which is 0 at both x=\pi and x=0.
Split into \dfrac{6x}{x^2+4}+\dfrac{3}{x^2+4}, with anti derivatives 3\log(x^2+4) and \dfrac32\tan^{-1}\left(\dfrac{x}{2}\right).
At x=2: 3\log8+\dfrac{3\pi}{8}. At x=0: 3\log4.
Anti derivatives: \displaystyle\int xe^x\,dx = xe^x-e^x (by parts), and \displaystyle\int \sin\dfrac{\pi x}{4}\,dx = -\dfrac{4}{\pi}\cos\dfrac{\pi x}{4}.
At x=1: -\dfrac{2\sqrt2}{\pi}. At x=0: -1-\dfrac{4}{\pi}.
Anti derivative \tan^{-1}x, evaluated as \tan^{-1}\sqrt3-\tan^{-1}1 = \dfrac{\pi}{3}-\dfrac{\pi}{4}.
Write 4+9x^2=9\left(x^2+\dfrac49\right), giving anti derivative \dfrac16\tan^{-1}\left(\dfrac{3x}{2}\right). Evaluate: \dfrac16\tan^{-1}(1)-0 = \dfrac16\cdot\dfrac{\pi}{4}.
Expert CBSE Coaching · Class 9–12