Q5. Find A(x) = ∫1x t2 dt, and check that A′(x) = x2 and A(1) = 0.
A(x) = [t3/3] from 1 to x = (x3 − 1)/3. Then A′(x) = x2, and A(1) = 0, as an area of zero width should be.
Q6. Beyond the NCERT text (JEE): find d/dx ∫0x3 et dt, first by the rule for a variable upper limit and then by integrating first.
By the rule: ex3 × 3x2. Integrating first: ∫0x3 et dt = ex3 − 1, whose derivative is 3x2ex3. The two agree.
Q7. Beyond the NCERT text (JEE): for F(x) = ∫0x (t2 − 4) dt, find the points where F′(x) = 0 and say which gives a local maximum.
By the first theorem F′(x) = x2 − 4, which is 0 at x = ±2. F″(x) = 2x is −4 < 0 at x = −2 (a local maximum) and 4 > 0 at x = 2 (a local minimum).
Q8. Why can Theorem 2 not be used for ∫02 dx/(x − 1)?
1/(x − 1) is not defined at x = 1, which lies in [0, 2], so the integrand is not continuous on the whole interval (NCERT’s Remark (iv), page 268). Writing [log |x − 1|] from 0 to 2 = 0 would be wrong.
Q9. Beyond the NCERT text (JEE): find d/dx of ∫xx2 et dt.
Both limits vary, so it is ex2 × 2x − ex. Check: the integral is ex2 − ex, whose derivative is the same.
Q10. Evaluate ∫1e (log x)/x dx and ∫02 x(2 − x)2 dx.
An anti derivative of (log x)/x is ½(log x)2, so the first is ½(1 − 0) = ½. For the second, expand to 4x − 4x2 + x3. Its anti derivative 2x2 − (4/3)x3 + x4/4 is 4/3 at 2 and 0 at 0, so the value is 4/3.